386
CHAPTER 14
Fluids
14-1 FLUIDS, DENSITY, AND PRESSURE
After reading this module, you should be able to . . .
14.01 Distinguish fluids from solids.
14.02 When mass is uniformly distributed, relate density to
mass and volume.
14.03 Apply the relationship between hydrostatic
pressure, force, and the surface area over which that
force acts.
●The density rof any material is defined as the material’s
mass per unit volume:
Usually, where a material sample is much larger than atomic
dimensions, we can write this as
●A fluid is a substance that can flow; it conforms to the
boundaries of its container because it cannot withstand
rm
V.
rm
V.
shearing stress. It can, however, exert a force perpendicular
to its surface. That force is described in terms of pressure p:
in which Fis the force acting on a surface element of
area A. If the force is uniform over a flat area, this can be
written as
●The force resulting from fluid pressure at a particular point
in a fluid has the same magnitude in all directions.
pF
A.
pF
A,
Learning Objectives
Key Ideas
What Is Physics?
The physics of fluids is the basis of hydraulic engineering,a branch of engineering
that is applied in a great many fields. A nuclear engineer might study the fluid
flow in the hydraulic system of an aging nuclear reactor, while a medical engineer
might study the blood flow in the arteries of an aging patient. An environmental
engineer might be concerned about the drainage from waste sites or the efficient
irrigation of farmlands. A naval engineer might be concerned with the dangers
faced by a deep-sea diver or with the possibility of a crew escaping from a
downed submarine.An aeronautical engineer might design the hydraulic systems
controlling the wing flaps that allow a jet airplane to land. Hydraulic engineering
is also applied in many Broadway and Las Vegas shows, where huge sets are
quickly put up and brought down by hydraulic systems.
Before we can study any such application of the physics of fluids, we must
first answer the question “What is a fluid?”
What Is a Fluid?
Afluid, in contrast to a solid, is a substance that can flow. Fluids conform to the
boundaries of any container in which we put them. They do so because a fluid
cannot sustain a force that is tangential to its surface. (In the more formal
language of Module 12-3, a fluid is a substance that flows because it cannot
387
14-1 FLUIDS, DENSITY, AND PRESSURE
withstand a shearing stress. It can, however, exert a force in the direction perpen-
dicular to its surface.) Some materials, such as pitch, take a long time to conform
to the boundaries of a container, but they do so eventually; thus, we classify even
those materials as fluids.
You may wonder why we lump liquids and gases together and call them fluids.
After all (you may say), liquid water is as different from steam as it is from ice.
Actually, it is not. Ice, like other crystalline solids, has its constituent atoms organ-
ized in a fairly rigid three-dimensional array called a crystalline lattice. In neither
steam nor liquid water,however,is there any such orderly long-range arrangement.
Density and Pressure
When we discuss rigid bodies, we are concerned with particular lumps of matter,
such as wooden blocks, baseballs, or metal rods. Physical quantities that we find
useful, and in whose terms we express Newton’s laws, are mass and force. We
might speak, for example,of a 3.6 kg block acted on by a 25 N force.
With fluids,we are more interested in the extended substance and in properties
that can vary from point to point in that substance. It is more useful to speak of
density and pressure than of mass and force.
Density
To find the density rof a fluid at any point, we isolate a small volume element V
around that point and measure the mass mof the fluid contained within that
element.The density is then
(14-1)
In theory, the density at any point in a fluid is the limit of this ratio as the volume
element Vat that point is made smaller and smaller. In practice, we assume that
a fluid sample is large relative to atomic dimensions and thus is “smooth” (with
uniform density), rather than “lumpy” with atoms. This assumption allows us to
write the density in terms of the mass mand volume Vof the sample:
(uniform density). (14-2)
Density is a scalar property; its SI unit is the kilogram per cubic meter.
Table 14-1 shows the densities of some substances and the average densities of
some objects. Note that the density of a gas (see Air in the table) varies consid-
erably with pressure, but the density of a liquid (see Water) does not; that is,
gases are readily compressible but liquids are not.
Pressure
Let a small pressure-sensing device be suspended inside a fluid-filled vessel, as
in Fig. 14-1a. The sensor (Fig. 14-1b) consists of a piston of surface area A
riding in a close-fitting cylinder and resting against a spring.A readout arrange-
ment allows us to record the amount by which the (calibrated) spring is
compressed by the surrounding fluid, thus indicating the magnitude Fof the
force that acts normal to the piston.We define the pressure on the piston as
(14-3)
In theory, the pressure at any point in the fluid is the limit of this ratio as the surface
area Aof the piston,centered on that point,is made smaller and smaller.However,
if the force is uniform over a flat area A(it is evenly distributed over every point of
pF
A.
rm
V
rm
V.
Table 14-1 Some Densities
Material or Object Density (kg/m3)
Interstellar space 1020
Best laboratory vacuum 1017
Air: 20C and 1 atm pressure 1.21
20C and 50 atm 60.5
Styrofoam 1 102
Ice 0.917 103
Water: 20C and 1 atm 0.998 103
20C and 50 atm 1.000 103
Seawater: 20C and 1 atm 1.024 103
Whole blood 1.060 103
Iron 7.9 103
Mercury (the metal,
not the planet) 13.6 103
Earth: average 5.5 103
core 9.5 103
crust 2.8 103
Sun: average 1.4 103
core 1.6 105
White dwarf star (core) 1010
Uranium nucleus 3 1017
Neutron star (core) 1018
Figure 14-1 (a) A fluid-filled vessel con-
taining a small pressure sensor,shown
in (b).The pressure is measured by the
relative position of the movable piston in
the sensor.
(a)
(b)
Pressure
sensor
Vacuum
Δ
ΔA
F
the area),we can write Eq.14-3 as
(pressure of uniform force on flat area), (14-4)
where Fis the magnitude of the normal force on area A.
We find by experiment that at a given point in a fluid at rest, the pressure p
defined by Eq. 14-4 has the same value no matter how the pressure sensor is
oriented. Pressure is a scalar, having no directional properties. It is true that
the force acting on the piston of our pressure sensor is a vector quantity, but
Eq. 14-4 involves only the magnitude of that force,a scalar quantity.
The SI unit of pressure is the newton per square meter, which is given a spe-
cial name, the pascal (Pa). In metric countries, tire pressure gauges are calibrated
in kilopascals. The pascal is related to some other common (non-SI) pressure
units as follows:
1 atm 1.01 105Pa 760 torr 14.7 lb/in.2.
The atmosphere (atm) is, as the name suggests, the approximate average pressure
of the atmosphere at sea level. The torr (named for Evangelista Torricelli, who
invented the mercury barometer in 1674) was formerly called the millimeter of
mercury (mm Hg).The pound per square inch is often abbreviated psi.Table 14-2
shows some pressures.
pF
A
388 CHAPTER 14 FLUIDS
Table 14-2 Some Pressures
Pressure (Pa)
Center of the Sun 2 1016
Center of Earth 4 1011
Highest sustained
laboratory pressure 1.5 1010
Deepest ocean trench
(bottom) 1.1 108
Spike heels on a dance floor 106
Automobile tirea2105
Atmosphere at sea level 1.0 105
Normal blood systolic
pressurea,b 1.6 104
Best laboratory vacuum 1012
aPressure in excess of atmospheric pressure.
bEquivalent to 120 torr on the physician’s
pressure gauge.
Sample Problem 14.01 Atmospheric pressure and force
A living room has floor dimensions of 3.5 m and 4.2 m and a
height of 2.4 m.
(a) What does the air in the room weigh when the air pres-
sure is 1.0 atm?
KEY IDEAS
(1) The air’s weight is equal to mg, where mis its mass.
(2) Mass mis related to the air density rand the air volume
Vby Eq. 14-2 (rm/V).
Calculation: Putting the two ideas together and taking the
density of air at 1.0 atm from Table 14-1, we find
mg (rV)g
(1.21 kg/m3)(3.5 m 4.2 m 2.4 m)(9.8 m/s2)
418 N 420 N. (Answer)
This is the weight of about 110 cans of Pepsi.
Additional examples, video, and practice available at WileyPLUS
14-2 FLUIDS AT REST
After reading this module, you should be able to . . .
14.04 Apply the relationship between the hydrostatic pressure,
fluid density, and the height above or below a reference level.
14.05 Distinguish between total pressure (absolute pressure)
and gauge pressure.
Learning Objectives
(b) What is the magnitude of the atmosphere’s downward
force on the top of your head, which we take to have an area
of 0.040 m2?
KEY IDEA
When the fluid pressure pon a surface of area Ais uniform,
the fluid force on the surface can be obtained from Eq. 14-4
(pF/A).
Calculation: Although air pressure varies daily, we can
approximate that p1.0 atm.Then Eq.14-4 gives
4.0 103N. (Answer)
This large force is equal to the weight of the air column from
the top of your head to the top of the atmosphere.
FpA (1.0 atm)
1.01 105N/m2
1.0 atm
(0.040 m2)
389
14-2 FLUIDS AT REST
y1
y2
y= 0
y
mg
y1
y2Level 2, p2
y= 0
y
F2
y1
y2
Level 1, p1
y= 0
y
F1
Sample
y1
y2
Air
Water
y= 0
y
(a)
Three forces act on this sample of water.
This upward force is due to the water
pressure pushing on the bottom surface. Gravity pulls downward on the sample.
Sample
mg
(e)F1
F2
The three forces
balance.
This downward force is due to the water
pressure pushing on the top surface.
(d)
(c)
(b)
A
Figure 14-2 (a) A tank of water in which a sample of water is contained in an imaginary cylinder of horizontal base area A.
(b)–(d) Force acts at the top surface of the cylinder;force acts at the bottom surface of the cylinder;the gravitational
force on the water in the cylinder is represented by . (e) A free-body diagram of the water sample. In WileyPLUS,this
figure is available as an animation with voiceover.
mg
:
F
:
2
F
:
1
●Pressure in a fluid at rest varies with vertical position y. For
ymeasured positive upward,
p2p1rg(y1y2).
If his the depth of a fluid sample below some reference level
at which the pressure is p0, this equation becomes
pp0rgh,
where pis the pressure in the sample.
●The pressure in a fluid is the same for all points at the same
level.
●Gauge pressure is the difference between the actual pres-
sure (or absolute pressure) at a point and the atmospheric
pressure.
Key Ideas
Fluids at Rest
Figure 14-2ashows a tank of water—or other liquid—open to the atmosphere.
As every diver knows, the pressure increases with depth below the air–water
interface. The diver’s depth gauge, in fact, is a pressure sensor much like that of
Fig. 14-1b. As every mountaineer knows, the pressure decreases with altitude as
one ascends into the atmosphere.The pressures encountered by the diver and the
mountaineer are usually called hydrostatic pressures, because they are due to flu-
ids that are static (at rest). Here we want to find an expression for hydrostatic
pressure as a function of depth or altitude.
Let us look first at the increase in pressure with depth below the water’s
surface. We set up a vertical yaxis in the tank, with its origin at the air–water
interface and the positive direction upward.We next consider a water sample con-
Figure 14-3 The pressure pincreases with
depth hbelow the liquid surface according
to Eq. 14-8.
p
h
Level 1
Level 2
Air
Liquid
y= 0
y
p0
The pressure at a point in a fluid in static equilibrium depends on the depth of
that point but not on any horizontal dimension of the fluid or its container.
390 CHAPTER 14 FLUIDS
tained in an imaginary right circular cylinder of horizontal base (or face) area A,
such that y1and y2(both of which are negative numbers) are the depths below the
surface of the upper and lower cylinder faces, respectively.
Figure 14-2eis a free-body diagram for the water in the cylinder.The water is
in static equilibrium; that is, it is stationary and the forces on it balance. Three
forces act on it vertically: Force acts at the top surface of the cylinder and is
due to the water above the cylinder (Fig. 14-2b). Force acts at the bottom sur-
face of the cylinder and is due to the water just below the cylinder (Fig. 14-2c).
The gravitational force on the water is m, where mis the mass of the water in the
cylinder (Fig. 14-2d).The balance of these forces is written as
F2F1mg. (14-5)
To involve pressures, we use Eq. 14-4 to write
F1p1Aand F2p2A. (14-6)
The mass mof the water in the cylinder is, from Eq. 14-2, mrV, where the
cylinder’s volume Vis the product of its face area Aand its height y1y2.Thus, m
is equal to rA(y1y2). Substituting this and Eq. 14-6 into Eq. 14-5, we find
p2Ap1ArAg(y1y2)
or p2p1rg(y1y2). (14-7)
This equation can be used to find pressure both in a liquid (as a function of
depth) and in the atmosphere (as a function of altitude or height). For the former,
suppose we seek the pressure pat a depth hbelow the liquid surface. Then we
choose level 1 to be the surface, level 2 to be a distance hbelow it (as in Fig. 14-3),
and p0to represent the atmospheric pressure on the surface.We then substitute
y10, p1p0and y2h,p2p
into Eq. 14-7, which becomes
pp0rgh (pressure at depth h). (14-8)
Note that the pressure at a given depth in the liquid depends on that depth but
not on any horizontal dimension.
g
:
F
:
2
F
:
1
Thus, Eq. 14-8 holds no matter what the shape of the container. If the bottom
surface of the container is at depth h, then Eq. 14-8 gives the pressure pthere.
In Eq. 14-8, pis said to be the total pressure, or absolute pressure, at level 2.
To see why, note in Fig. 14-3 that the pressure pat level 2 consists of two contribu-
tions: (1) p0, the pressure due to the atmosphere, which bears down on the liquid,
and (2) rgh, the pressure due to the liquid above level 2, which bears down on
level 2. In general, the difference between an absolute pressure and an atmo-
spheric pressure is called the gauge pressure (because we use a gauge to measure
this pressure difference). For Fig.14-3, the gauge pressure is rgh.
Equation 14-7 also holds above the liquid surface: It gives the atmospheric pres-
sure at a given distance above level 1 in terms of the atmospheric pressure p1at level 1
(assuming that the atmospheric density is uniform over that distance). For example, to
find the atmospheric pressure at a distance dabove level 1 in Fig.14-3,we substitute
y10, p1p0and y2d,p2p.
Then with rrair, we obtain
pp0rairgd.
391
14-2 FLUIDS AT REST
Checkpoint 1
The figure shows four
containers of olive oil.
Rank them according
to the pressure at
depth h, greatest first.
h
(
a
)(
b
)(
c
)(
d
)
ascends, the external pressure on him decreases, until it is
atmospheric pressure p0at the surface. His blood pressure
also decreases, until it is normal. However, because he does
not exhale, the air pressure in his lungs remains at the value it
had at depth L. At the surface, the pressure difference pis
ppp0rgL,
0.95 m. (Answer)
This is not deep! Yet, the pressure difference of 9.3 kPa
(about 9% of atmospheric pressure) is sufficient to rupture
the diver’s lungs and force air from them into the depres-
surized blood, which then carries the air to the heart,
killing the diver. If the diver follows instructions and grad-
ually exhales as he ascends, he allows the pressure in his
lungs to equalize with the external pressure, and then there
is no danger.
Lp
g9300 Pa
(998 kg/m3)(9.8 m/s2)
Sample Problem 14.02 Gauge pressure on a scuba diver
A novice scuba diver practicing in a swimming pool takes
enough air from his tank to fully expand his lungs before
abandoning the tank at depth Land swimming to the sur-
face, failing to exhale during his ascent. At the surface, the
difference pbetween the external pressure on him and the
air pressure in his lungs is 9.3 kPa. From what depth does he
start? What potentially lethal danger does he face?
KEY IDEA
The pressure at depth hin a liquid of density ris given by
Eq. 14-8 (pp0rgh), where the gauge pressure rgh is
added to the atmospheric pressure p0.
Calculations: Here, when the diver fills his lungs at depth L,
the external pressure on him (and thus the air pressure within
his lungs) is greater than normal and given by Eq. 14-8 as
pp0rgL,
where ris the water’s density (998 kg/m3,Table 14-1). As he
Equating these two expressions and solving for the un-
known density yield
915 kg/m3. (Answer)
Note that the answer does not depend on the atmospheric
pressure p0or the free-fall acceleration g.
rxrw
l
ld(998 kg/m3)135 mm
135 mm 12.3 mm
The U-tube in Fig. 14-4 contains two liquids in static equilib-
rium: Water of density rw(998 kg/m3) is in the right arm,
and oil of unknown density rxis in the left.Measurement gives
l135 mm and d12.3 mm. What is the density of the oil?
KEY IDEAS
(1) The pressure pint at the level of the oil–water interface in
the left arm depends on the density rxand height of the oil
above the interface. (2) The water in the right arm at the
same level must be at the same pressure pint. The reason is
that, because the water is in static equilibrium, pressures at
points in the water at the same level must be the same.
Calculations: In the right arm, the interface is a distance l
below the free surface of the water, and we have, from Eq. 14-8,
pint p0rwgl (right arm).
In the left arm, the interface is a distance ldbelow the free
surface of the oil, and we have,again from Eq.14-8,
pint p0rxg(ld)(left arm).
Additional examples, video, and practice available at WileyPLUS
Figure 14-4 The oil in the left arm stands higher than the water.
Interface
Water
Oil
l
d
This much oil
balances... ... this much
water.
Sample Problem 14.03 Balancing of pressure in a U-tube
so
392 CHAPTER 14 FLUIDS
Figure 14-6 An open-tube manometer, con-
nected to measure the gauge pressure of
the gas in the tank on the left.The right arm
of the U-tube is open to the atmosphere.
Tank
M
a
n
o
m
ete
r
Level 2
Level 1
p
0
h
pg
Measuring Pressure
The Mercury Barometer
Figure 14-5ashows a very basic mercury barometer, a device used to
measure the pressure of the atmosphere. The long glass tube is filled
with mercury and inverted with its open end in a dish of mercury, as
the figure shows. The space above the mercury column contains only
mercury vapor, whose pressure is so small at ordinary temperatures
that it can be neglected.
We can use Eq. 14-7 to find the atmospheric pressure p0in terms
of the height hof the mercury column.We choose level 1 of Fig. 14-2 to
be that of the air–mercury interface and level 2 to be that of the top of
the mercury column, as labeled in Fig.14-5a.We then substitute
y10, p1p0and y2h,p20
into Eq. 14-7, finding that
p0rgh, (14-9)
where ris the density of the mercury.
For a given pressure,the height hof the mercury column does not
depend on the cross-sectional area of the vertical tube. The fanciful
mercury barometer of Fig. 14-5bgives the same reading as that of Fig. 14-5a; all
that counts is the vertical distance hbetween the mercury levels.
Equation 14-9 shows that, for a given pressure, the height of the column of
mercury depends on the value of gat the location of the barometer and on the
density of mercury, which varies with temperature. The height of the column (in
millimeters) is numerically equal to the pressure (in torr) only if the barometer is
at a place where ghas its accepted standard value of 9.80665 m/s2and the
temperature of the mercury is 0°C. If these conditions do not prevail (and they
rarely do), small corrections must be made before the height of the mercury
column can be transformed into a pressure.
The Open-Tube Manometer
An open-tube manometer (Fig. 14-6) measures the gauge pressure pgof a gas. It
consists of a U-tube containing a liquid, with one end of the tube connected to the
vessel whose gauge pressure we wish to measure and the other end open to the
atmosphere.We can use Eq. 14-7 to find the gauge pressure in terms of the height
hshown in Fig.14-6. Let us choose levels 1 and 2 as shown in Fig. 14-6.With
y10, p1p0and y2h,p2p
substituted into Eq. 14-7, we find that
pgpp0rgh, (14-10)
where ris the liquid’s density. The gauge pressure pgis directly proportional to h.
14-3 MEASURING PRESSURE
After reading this module, you should be able to . . .
14.06 Describe how a barometer can measure atmospheric
pressure.
14.07 Describe how an open-tube manometer can measure
the gauge pressure of a gas.
Learning Objectives
●A mercury barometer can be used to measure atmospheric
pressure.
●An open-tube manometer can be used to measure the
gauge pressure of a confined gas.
Key Ideas
Level 1
p0
y
Level 2
h
p≈ 0
h
p0
p≈ 0
(a) (b)
Figure 14-5 (a) A mercury barometer. (b)
Another mercury barometer.The distance
his the same in both cases.
393
The gauge pressure can be positive or negative, depending on whether
pp0or pp0. In inflated tires or the human circulatory system, the
(absolute) pressure is greater than atmospheric pressure, so the gauge pressure is a
positive quantity, sometimes called the overpressure. If you suck on a straw to pull
fluid up the straw, the (absolute) pressure in your lungs is actually less than atmo-
spheric pressure.The gauge pressure in your lungs is then a negative quantity.
14-4 PASCAL’S PRINCIPLE
14-4 PASCAL’S PRINCIPLE
After reading this module, you should be able to . . .
14.08 Identify Pascal’s principle.
14.09 For a hydraulic lift, apply the relationship between the
input area and displacement and the output area and
displacement.
Learning Objectives
●Pascal’s principle states that a change in the pressure applied to an enclosed fluid is transmitted undiminished to every por-
tion of the fluid and to the walls of the containing vessel.
Key Idea
Pascal’s Principle
When you squeeze one end of a tube to get toothpaste out the other end, you are
watching Pascal’s principle in action.This principle is also the basis for the Heimlich
maneuver, in which a sharp pressure increase properly applied to the abdomen is
transmitted to the throat, forcefully ejecting food lodged there. The principle was
first stated clearly in 1652 by Blaise Pascal (for whom the unit of pressure is named):
Figure 14-7 Lead shot (small balls of lead)
loaded onto the piston create a pressure pext
at the top of the enclosed (incompressible)
liquid. If pext is increased, by adding more
lead shot, the pressure increases by the same
amount at all points within the liquid.
Lead shot
Piston
Pp
h
pext
Liquid
A change in the pressure applied to an enclosed incompressible fluid is transmit-
ted undiminished to every portion of the fluid and to the walls of its container.
Demonstrating Pascal’s Principle
Consider the case in which the incompressible fluid is a liquid contained in a tall
cylinder, as in Fig. 14-7.The cylinder is fitted with a piston on which a container of
lead shot rests.The atmosphere, container, and shot exert pressure pext on the pis-
ton and thus on the liquid.The pressure pat any point Pin the liquid is then
ppext rgh. (14-11)
Let us add a little more lead shot to the container to increase pext by an amount
pext. The quantities r,g, and hin Eq. 14-11 are unchanged, so the pressure
change at Pis
ppext. (14-12)
This pressure change is independent of h, so it must hold for all points within the
liquid, as Pascal’s principle states.
Pascal’s Principle and the Hydraulic Lever
Figure 14-8 shows how Pascal’s principle can be made the basis of a hydraulic lever.
In operation,let an external force of magnitude Fibe directed downward on the left-
hand (or input) piston, whose surface area is Ai.An incompressible liquid in the de-
vice then produces an upward force of magnitude Foon the right-hand (or output)
piston, whose surface area is Ao.To keep the system in equilibrium, there must be a
downward force of magnitude Foon the output piston from an external load (not
di
Input
Ai
do
Oil
Ao
Output
Fi
Fo
A small input
force produces ...
... a large output
force.
Figure 14-8 A hydraulic arrangement that
can be used to magnify a force .The work
done is,however,not magnified and is the
same for both the input and output forces.
F
:
i
394 CHAPTER 14 FLUIDS
14-5 ARCHIMEDES’ PRINCIPLE
After reading this module, you should be able to . . .
14.10 Describe Archimedes’ principle.
14.11 Apply the relationship between the buoyant force on a
body and the mass of the fluid displaced by the body.
14.12 For a floating body, relate the buoyant force to the
gravitational force.
14.13 For a floating body, relate the gravitational force to the
mass of the fluid displaced by the body.
14.14 Distinguish between apparent weight and actual weight.
14.15 Calculate the apparent weight of a body that is fully or
partially submerged.
Learning Objectives
●Archimedes’ principle states that when a body is fully or
partially submerged in a fluid, the fluid pushes upward with a
buoyant force with magnitude
Fbmfg,
where mfis the mass of the fluid that has been pushed out of
the way by the body.
●When a body floats in a fluid, the magnitude Fbof the
(upward) buoyant force on the body is equal to the magnitude
Fgof the (downward) gravitational force on the body.
●The apparent weight of a body on which a buoyant force
acts is related to its actual weight by
weightapp weight Fb.
Key Ideas
shown).The force applied on the left and the downward force from the load onF
:
o
F
:
i
With a hydraulic lever, a given force applied over a given distance can be
transformed to a greater force applied over a smaller distance.
the right produce a change pin the pressure of the liquid that is given by
,
so . (14-13)
Equation 14-13 shows that the output force Foon the load must be greater than
the input force Fiif AoAi, as is the case in Fig. 14-8.
If we move the input piston downward a distance di, the output piston moves
upward a distance do, such that the same volume Vof the incompressible liquid is
displaced at both pistons.Then
VAidiAodo,
which we can write as
. (14-14)
This shows that, if AoAi(as in Fig. 14-8), the output piston moves a smaller
distance than the input piston moves.
From Eqs. 14-13 and 14-14 we can write the output work as
(14-15)
which shows that the work Wdone on the input piston by the applied force is
equal to the work Wdone by the output piston in lifting the load placed on it.
The advantage of a hydraulic lever is this:
WF
odo
F
i
Ao
Ai
di
Ai
Ao
F
idi,
dodi
Ai
Ao
F
oF
i
Ao
Ai
pF
i
Ai
F
o
Ao
The product of force and distance remains unchanged so that the same work is
done. However, there is often tremendous advantage in being able to exert the
larger force. Most of us, for example, cannot lift an automobile directly but can
with a hydraulic jack, even though we have to pump the handle farther than
the automobile rises and in a series of small strokes.
395
14-5 ARCHIMEDES’ PRINCIPLE
The upward buoyant
force on this sack of
water equals the
weight of the water.
Figure 14-9 A thin-walled plastic sack of water
is in static equilibrium in the pool.The gravita-
tional force on the sack must be balanced by
a net upward force on it from the surrounding
water.
(a)(b)
FbThe buoyant force
is due to the
pressure of the
surrounding water.
Stone
Fb
Fg
The net force is
downward, so the
stone accelerates
downward.
(c)
Wood
Fb
Fg
The net force
is upward, so the
wood accelerates
upward.
Figure 14-10 (a) The water surrounding the hole in the water pro-
duces a net upward buoyant force on whatever fills the hole.
(b) For a stone of the same volume as the hole,the gravita-
tional force exceeds the buoyant force in magnitude. (c) For a
lump of wood of the same volume, the gravitational force is
less than the buoyant force in magnitude.
the magnitude mfgof the gravitational force on the sack of water: Fb mfg.F
:
g
(Subscript frefers to fluid, here the water.) In words, the magnitude of the buoyant
force is equal to the weight of the water in the sack.
In Fig. 14-10b, we have replaced the sack of water with a stone that exactly fills
the hole in Fig. 14-10a.The stone is said to displace the water, meaning that the stone
occupies space that would otherwise be occupied by water.We have changed nothing
about the shape of the hole, so the forces at the hole’s surface must be the same as
when the water-filled sack was in place. Thus, the same upward buoyant force that
acted on the water-filled sack now acts on the stone; that is, the magnitude Fbof the
buoyant force is equal to mfg,the weight of the water displaced by the stone.
Unlike the water-filled sack, the stone is not in static equilibrium. The down-
ward gravitational force on the stone is greater in magnitude than the upward
buoyant force (Fig.14-10b).The stone thus accelerates downward,sinking.
Let us next exactly fill the hole in Fig. 14-10awith a block of lightweight
wood, as in Fig. 14-10c.Again, nothing has changed about the forces at the hole’s
surface, so the magnitude Fbof the buoyant force is still equal to mfg, the weight
F
:
g
Archimedes’ Principle
Figure 14-9 shows a student in a swimming pool, manipulating a very thin plastic
sack (of negligible mass) that is filled with water. She finds that the sack and its
contained water are in static equilibrium, tending neither to rise nor to sink.
The downward gravitational force on the contained water must be balanced
by a net upward force from the water surrounding the sack.
This net upward force is a buoyant force . It exists because the pressure in
the surrounding water increases with depth below the surface. Thus, the pressure
near the bottom of the sack is greater than the pressure near the top, which
means the forces on the sack due to this pressure are greater in magnitude near
the bottom of the sack than near the top. Some of the forces are represented in
Fig. 14-10a, where the space occupied by the sack has been left empty. Note that the
force vectors drawn near the bottom of that space (with upward components) have
longer lengths than those drawn near the top of the sack (with downward compo-
nents). If we vectorially add all the forces on the sack from the water, the horizontal
components cancel and the vertical components add to yield the upward buoyant
force on the sack.(Force is shown to the right of the pool in Fig.14-10a.)
Because the sack of water is in static equilibrium, the magnitude of is equal toF
:
b
F
:
b
F
:
b
F
:
b
F
:
g
396 CHAPTER 14 FLUIDS
When a body floats in a fluid, the magnitude Fbof the buoyant force on the body
is equal to the magnitude Fgof the gravitational force on the body.
When a body floats in a fluid, the magnitude Fgof the gravitational force on the
body is equal to the weight mfgof the fluid that has been displaced by the body.
of the displaced water. Like the stone, the block is not in static equilibrium.
However, this time the gravitational force is lesser in magnitude than the
buoyant force (as shown to the right of the pool), and so the block accelerates
upward, rising to the top surface of the water.
Our results with the sack, stone, and block apply to all fluids and are summarized
in Archimedes’ principle:
F
:
g
When a body is fully or partially submerged in a fluid, a buoyant force from the
surrounding fluid acts on the body. The force is directed upward and has a magni-
tude equal to the weight mfgof the fluid that has been displaced by the body.
F
:
b
The buoyant force on a body in a fluid has the magnitude
Fbmfg(buoyant force), (14-16)
where mfis the mass of the fluid that is displaced by the body.
Floating
When we release a block of lightweight wood just above the water in a pool,the block
moves into the water because the gravitational force on it pulls it downward. As the
block displaces more and more water, the magnitude Fbof the upward buoyant force
acting on it increases. Eventually, Fbis large enough to equal the magnitude Fgof the
downward gravitational force on the block, and the block comes to rest. The block is
then in static equilibrium and is said to be floating in the water.In general,
We can write this statement as
FbFg(floating). (14-17)
From Eq. 14-16, we know that Fbmfg. Thus,
We can write this statement as
Fgmfg(floating). (14-18)
In other words, a floating body displaces its own weight of fluid.
Apparent Weight in a Fluid
If we place a stone on a scale that is calibrated to measure weight, then the
reading on the scale is the stone’s weight. However, if we do this underwater,
the upward buoyant force on the stone from the water decreases the reading.
That reading is then an apparent weight. In general, an apparent weight is related
to the actual weight of a body and the buoyant force on the body by
which we can write as
weightapp weight Fb(apparent weight). (14-19)
apparent
weight
actual
weight
magnitude of
buoyant force
,
397
If, in some test of strength, you had to lift a heavy stone, you could do it more
easily with the stone underwater.Then your applied force would need to exceed
only the stone’s apparent weight,not its larger actual weight.
The magnitude of the buoyant force on a floating body is equal to the body’s
weight. Equation 14-19 thus tells us that a floating body has an apparent weight of
zero—the body would produce a reading of zero on a scale. For example, when as-
tronauts prepare to perform a complex task in space, they practice the task floating
underwater, where their suits are adjusted to give them an apparent weight of zero.
14-5 ARCHIMEDES’ PRINCIPLE
Checkpoint 2
A penguin floats first in a fluid of density r0, then in a fluid of density 0.95r0, and then
in a fluid of density 1.1r0. (a) Rank the densities according to the magnitude of the
buoyant force on the penguin, greatest first. (b) Rank the densities according to the
amount of fluid displaced by the penguin, greatest first.
Sample Problem 14.04 Floating, buoyancy, and density
In Fig. 14-11, a block of density floats face
down in a fluid of density . The block has
height .
(a) By what depth his the block submerged?
KEY IDEAS
(1) Floating requires that the upward buoyant force on the
block match the downward gravitational force on the block.
(2) The buoyant force is equal to the weight of the fluid
displaced by the submerged portion of the block.
Calculations: From Eq. 14-16, we know that the buoyant
force has the magnitude , where is the mass of
the fluid displaced by the block’s submerged volume
From Eq. 14-2 , we know that the mass of the dis-
placed fluid is We don’t know but if we symbol-
ize the block’s face length as Land its width as W, then from
Fig. 14-11 we see that the submerged volume must be
. If we now combine our three expressions, we
find that the upward buoyant force has magnitude
(14-20)
Similarly, we can write the magnitude of the gravita-
tional force on the block, first in terms of the block’s mass
m, then in terms of the block’s density rand (full) volume V,
and then in terms of the block’s dimensions L,W, and H
(the full height):
. (14-21)
The floating block is stationary. Thus, writing Newton’s
second law for components along a vertical yaxis with the
positive direction upward , we have
FbFgm(0),
(F
net,ymay)
Fgmg rVg rfLWHg
Fg
FbmfgrfVfgrfLWhg.
V
fLWh
V
f
mfrfV
f.
(rm/V)
Vf.
mf
Fbmfg
mfg
H6.0 cm
rf1200 kg/m3
r800 kg/m3
or from Eqs. 14-20 and 14-21,
which gives us
. (Answer)
(b) If the block is held fully submerged and then released,
what is the magnitude of its acceleration?
Calculations: The gravitational force on the block is the
same but now, with the block fully submerged, the volume
of the displaced water is (The full height of
the block is used.) This means that the value of is now
larger, and the block will no longer be stationary but will
accelerate upward. Now Newton’s second law yields
,
or ,
where we inserted for the mass mof the block. Solv-
ing for aleads to
(Answer)4.9 m/s2.
a
rf
r1
g
1200 kg/m3
800 kg/m31
(9.8 m/s2)
rLWH
rfLWHg rLWHg rLWHa
FbFgma
Fb
VLWH.
4.0 cm
hr
rfH800 kg/m3
1200 kg/m3 (6.0 cm)
rfLWhg rLWHg 0,
h
H
Floating means
that the buoyant
force matches the
gravitational force.
Figure 14-11 Block of height Hfloats
in a fluid, to a depth of h.
Additional examples, video, and practice available at WileyPLUS
Floating means
that the buoyant
force matches the
gravitational force.
398 CHAPTER 14 FLUIDS
Ideal Fluids in Motion
The motion of real fluids is very complicated and not yet fully understood.
Instead, we shall discuss the motion of an ideal fluid, which is simpler to handle
mathematically and yet provides useful results. Here are four assumptions that
we make about our ideal fluid; they all are concerned with flow:
1. Steady flow In steady (or laminar)flow, the velocity of the moving fluid at any
fixed point does not change with time. The gentle flow of water near the center
of a quiet stream is steady; the flow in a chain of rapids is not. Figure 14-12 shows
a transition from steady flow to nonsteady (or nonlaminar or turbulent)flow
for a rising stream of smoke. The speed of the smoke particles increases as
they rise and, at a certain critical speed, the flow changes from steady to non-
steady.
2. Incompressible flow We assume, as for fluids at rest, that our ideal fluid is
incompressible; that is, its density has a constant, uniform value.
3. Nonviscous flow Roughly speaking, the viscosity of a fluid is a measure of how
resistive the fluid is to flow. For example, thick honey is more resistive to flow than
water, and so honey is said to be more viscous than water. Viscosity is the fluid
analog of friction between solids; both are mechanisms by which the kinetic en-
ergy of moving objects can be transferred to thermal energy. In the absence of fric-
tion, a block could glide at constant speed along a horizontal surface. In the same
way, an object moving through a nonviscous fluid would experience no viscous
drag force —that is, no resistive force due to viscosity; it could move at constant
speed through the fluid.The British scientist Lord Rayleigh noted that in an ideal
fluid a ship’s propeller would not work, but, on the other hand, in an ideal fluid a
ship (once set into motion) would not need a propeller!
4. Irrotational flow Although it need not concern us further, we also assume
that the flow is irrotational. To test for this property, let a tiny grain of dust
move with the fluid.Although this test body may (or may not) move in a circu-
lar path, in irrotational flow the test body will not rotate about an axis through
its own center of mass. For a loose analogy, the motion of a Ferris wheel is ro-
tational; that of its passengers is irrotational.
We can make the flow of a fluid visible by adding a tracer. This might
be a dye injected into many points across a liquid stream (Fig. 14-13) or smoke
●An ideal fluid is incompressible and lacks viscosity, and its
flow is steady and irrotational.
●Astreamline is the path followed by an individual fluid particle.
●Atube of flow is a bundle of streamlines.
●The flow within any tube of flow obeys the equation of continuity:
RVAv a constant,
in which RVis the volume flow rate, Ais the cross-sectional
area of the tube of flow at any point, and vis the speed of the
fluid at that point.
●The mass flow rate Rmis
RmrRVrAv a constant.
Key Ideas
Figure 14-12 At a certain point, the rising flow
of smoke and heated gas changes from
steady to turbulent.
Will McIntyre/Photo Researchers, Inc.
14-6 THE EQUATION OF CONTINUITY
After reading this module, you should be able to . . .
14.16 Describe steady flow, incompressible flow, nonviscous
flow, and irrotational flow.
14.17 Explain the term streamline.
14.18 Apply the equation of continuity to relate the
cross-sectional area and flow speed at one point in a tube
to those quantities at a different point.
14.19 Identify and calculate volume flow rate.
14.20 Identify and calculate mass flow rate.
Learning Objectives
399
14-6 THE EQUATION OF CONTINUITY
particles added to a gas flow (Fig. 14-12). Each bit of a tracer follows a stream-
line, which is the path that a tiny element of the fluid would take as the fluid
flows. Recall from Chapter 4 that the velocity of a particle is always tangent to
the path taken by the particle. Here the particle is the fluid element, and its ve-
locity is always tangent to a streamline (Fig. 14-14). For this reason, two
streamlines can never intersect; if they did, then an element arriving at
their intersection would have two different velocities simultaneously—an
impossibility.
The Equation of Continuity
You may have noticed that you can increase the speed of the water emerging
from a garden hose by partially closing the hose opening with your thumb.
Apparently the speed vof the water depends on the cross-sectional area A
through which the water flows.
Here we wish to derive an expression that relates vand Afor the steady flow
of an ideal fluid through a tube with varying cross section, like that in Fig. 14-15.
The flow there is toward the right, and the tube segment shown (part of a longer
tube) has length L.The fluid has speeds v1at the left end of the segment and v2at
the right end. The tube has cross-sectional areas A1at the left end and A2at the
right end. Suppose that in a time interval ta volume Vof fluid enters the tube
segment at its left end (that volume is colored purple in Fig. 14-15).Then, because
the fluid is incompressible, an identical volume Vmust emerge from the right
end of the segment (it is colored green in Fig.14-15).
v
:
Courtesy D. H. Peregrine, University of Bristol
Figure 14-13 The steady flow
of a fluid around a cylin-
der,as revealed by a dye
tracer that was injected
into the fluid upstream of
the cylinder.
Streamline
Fluid
e
l
e
m
e
n
t
v
Figure 14-14 A fluid element traces out a
streamline as it moves.The velocity vector
of the element is tangent to the streamline
at every point.
Figure 14-15 Fluid flows from left to right at a steady
rate through a tube segment of length L. The fluid’s
speed is v1at the left side and v2at the right side.The
tube’s cross-sectional area is A1at the left side and
A2at the right side. From time tin (a) to time tt
in (b), the amount of fluid shown in purple enters at
the left side and the equal amount of fluid shown in
green emerges at the right side.
L
v1
A1
A2
v2
(a) Time t
L
(b) Time t + Δt
The volume flow per
second here must
match ...
... the volume flow
per second here.
400 CHAPTER 14 FLUIDS
Figure 14-16 Fluid flows at a constant speed v
through a tube. (a) At time t, fluid element e
is about to pass the dashed line. (b) At time
tt, element eis a distance xvt
from the dashed line.
v
e
v
e
(a) Time t
(b) Time t + Δt
Δ x
A1
A2
The volume
flow per
second here
must match ...
... the volume flow
per second here.
Figure 14-17 A tube of flow is defined by the
streamlines that form the boundary of the
tube.The volume flow rate must be the same
for all cross sections of the tube of flow.
Checkpoint 3
The figure shows a pipe and
gives the volume flow rate
(in cm3/s) and the direction of
flow for all but one section.
What are the volume flow
rate and the direction of flow
for that section?
4 8
2 5 6
4
We can use this common volume Vto relate the speeds and areas. To do
so, we first consider Fig. 14-16, which shows a side view of a tube of uniform
cross-sectional area A. In Fig. 14-16a, a fluid element eis about to pass through
the dashed line drawn across the tube width. The element’s speed is v, so dur-
ing a time interval t, the element moves along the tube a distance xvt.
The volume Vof fluid that has passed through the dashed line in that time
interval tis
VAxAv t. (14-22)
Applying Eq. 14-22 to both the left and right ends of the tube segment in
Fig.14-15, we have
VA1v1tA2v2t
or A1v1A2v2(equation of continuity). (14-23)
This relation between speed and cross-sectional area is called the equation of
continuity for the flow of an ideal fluid. It tells us that the flow speed increases
when we decrease the cross-sectional area through which the fluid flows.
Equation 14-23 applies not only to an actual tube but also to any so-called
tube of flow, or imaginary tube whose boundary consists of streamlines. Such
a tube acts like a real tube because no fluid element can cross a streamline;
thus, all the fluid within a tube of flow must remain within its boundary.
Figure 14-17 shows a tube of flow in which the cross-sectional area increases
from area A1to area A2along the flow direction. From Eq. 14-23 we know
that, with the increase in area, the speed must decrease, as is indicated by the
greater spacing between streamlines at the right in Fig. 14-17. Similarly, you
can see that in Fig. 14-13 the speed of the flow is greatest just above and just
below the cylinder.
We can rewrite Eq. 14-23 as
RVAv a constant (volume flow rate,equation of continuity), (14-24)
in which RVis the volume flow rate of the fluid (volume past a given point per
unit time). Its SI unit is the cubic meter per second (m3/s). If the density rof the
fluid is uniform, we can multiply Eq. 14-24 by that density to get the mass flow
rate Rm(mass per unit time):
RmrRVrAv a constant (mass flow rate). (14-25)
The SI unit of mass flow rate is the kilogram per second (kg/s). Equation 14-25
says that the mass that flows into the tube segment of Fig. 14-15 each second must
be equal to the mass that flows out of that segment each second.
401
14-7 BERNOULLI’S EQUATION
Figure 14-18 As water falls from a tap,its speed in-
creases. Because the volume flow rate must be the
same at all horizontal cross sections of the stream,
the stream must “neck down” (narrow).
h
A0
A
The volume flow per
second here must
match ...
... the volume flow
per second here.
KEY IDEA
The volume flow rate through the higher cross section must
be the same as that through the lower cross section.
Calculations: From Eq. 14-24, we have
A0v0Av, (14-26)
where v0and vare the water speeds at the levels correspon-
ding to A0and A. From Eq. 2-16 we can also write, because
the water is falling freely with acceleration g,
(14-27)
Eliminating vbetween Eqs. 14-26 and 14-27 and solving for
v0, we obtain
0.286 m/s 28.6 cm/s.
From Eq. 14-24, the volume flow rate RVis then
RVA0v0(1.2 cm2)(28.6 cm/s)
34 cm3/s. (Answer)
A(2)(9.8 m/s2)(0.045 m)(0.35 cm2)2
(1.2 cm2)2(0.35 cm2)2
v0A2ghA2
A2
0A2
v2v2
02gh.
Sample Problem 14.05 A water stream narrows as it falls
Figure 14-18 shows how the stream of water emerging from a
faucet “necks down” as it falls. This change in the horizontal
cross-sectional area is characteristic of any laminar (non-
turbulant) falling stream because the gravitational force
increases the speed of the stream. Here the indicated cross-
sectional areas are A01.2 cm2and A0.35 cm2. The two
levels are separated by a vertical distance h45 mm. What
is the volume flow rate from the tap?
Additional examples, video, and practice available at WileyPLUS
14-7 BERNOULLI’S EQUATION
After reading this module, you should be able to . . .
14.21 Calculate the kinetic energy density in terms of a fluid’s
density and flow speed.
14.22 Identify the fluid pressure as being a type of energy
density.
14.23 Calculate the gravitational potential energy density.
14.24 Apply Bernoulli’s equation to relate the total energy
density at one point on a streamline to the value at another
point.
14.25 Identify that Bernoulli's equation is a statement of the
conservation of energy.
Learning Objectives
●Applying the principle of conservation of mechanical energy to the flow of an ideal fluid leads to Bernoulli’s equation:
prv2rgy a constant
along any tube of flow.
1
2
Key Idea
Bernoulli’s Equation
Figure 14-19 represents a tube through which an ideal fluid is flowing at a steady
rate. In a time interval t, suppose that a volume of fluid V, colored purple in
Fig. 14-19, enters the tube at the left (or input) end and an identical volume,
402 CHAPTER 14 FLUIDS
Figure 14-19 Fluid flows at a steady rate
through a length Lof a tube, from the
input end at the left to the output end at
the right. From time tin (a) to time tt
in (b), the amount of fluid shown in
purple enters the input end and the
equal amount shown in green emerges
from the output end.
p1
L
Input
v1
y1
(a)
(b)
y
v2
p2
y2
y
x
t
t+Δt
x
Output
*For irrotational flow (which we assume), the constant in Eq. 14-29 has the same value for all
points within the tube of flow; the points do not have to lie along the same streamline. Similarly,
the points 1 and 2 in Eq. 14-28 can lie anywhere within the tube of flow.
If the speed of a fluid element increases as the element travels along a horizontal
streamline, the pressure of the fluid must decrease, and conversely.
colored green in Fig. 14-19, emerges at the right (or output) end. The emerging
volume must be the same as the entering volume because the fluid is incompress-
ible, with an assumed constant density r.
Let y1,v1, and p1be the elevation, speed, and pressure of the fluid entering at
the left, and y2,v2, and p2be the corresponding quantities for the fluid emerging
at the right. By applying the principle of conservation of energy to the fluid, we
shall show that these quantities are related by
(14-28)
In general, the term is called the fluid’s kinetic energy density (kinetic en-
ergy per unit volume).We can also write Eq. 14-28 as
(Bernoulli’s equation). (14-29)
Equations 14-28 and 14-29 are equivalent forms of Bernoulli’s equation,
after Daniel Bernoulli, who studied fluid flow in the 1700s.* Like the equation of
continuity (Eq. 14-24), Bernoulli’s equation is not a new principle but simply
the reformulation of a familiar principle in a form more suitable to fluid
mechanics. As a check, let us apply Bernoulli’s equation to fluids at rest, by put-
ting v1v20 in Eq. 14-28.The result is Eq. 14-7:
p2p1rg(y1y2).
A major prediction of Bernoulli’s equation emerges if we take yto be a
constant (y0, say) so that the fluid does not change elevation as it flows. Equation
14-28 then becomes
(14-30)
which tells us that:
p11
2rv2
1p21
2rv2
2,
p1
2rv2rgy a constant
1
2rv2
p11
2rv2
1rgy1p21
2rv2
2rgy2.
Put another way, where the streamlines are relatively close together (where the
velocity is relatively great), the pressure is relatively low, and conversely.
The link between a change in speed and a change in pressure makes sense
if you consider a fluid element that travels through a tube of various widths.
Recall that the element’s speed in the narrower regions is fast and its speed in the
wider regions is slow. By Newton’s second law, forces (or pressures) must cause
the changes in speed (the accelerations). When the element nears a narrow re-
gion, the higher pressure behind it accelerates it so that it then has a greater
speed in the narrow region. When it nears a wide region, the higher pressure
ahead of it decelerates it so that it then has a lesser speed in the wide region.
Bernoulli’s equation is strictly valid only to the extent that the fluid is ideal. If
viscous forces are present, thermal energy will be involved, which here we neglect.
Proof of Bernoulli’s Equation
Let us take as our system the entire volume of the (ideal) fluid shown in
Fig. 14-19.We shall apply the principle of conservation of energy to this system as
it moves from its initial state (Fig. 14-19a) to its final state (Fig. 14-19b). The fluid
lying between the two vertical planes separated by a distance Lin Fig. 14-19 does
not change its properties during this process; we need be concerned only with
changes that take place at the input and output ends.
403
14-7 BERNOULLI’S EQUATION
First, we apply energy conservation in the form of the work–kinetic energy
theorem,
WK, (14-31)
which tells us that the change in the kinetic energy of our system must equal the
net work done on the system. The change in kinetic energy results from the
change in speed between the ends of the tube and is
, (14-32)
in which m(rV) is the mass of the fluid that enters at the input end and
leaves at the output end during a small time interval t.
The work done on the system arises from two sources. The work Wgdone by
the gravitational force on the fluid of mass mduring the vertical lift of
the mass from the input level to the output level is
Wgmg(y2y1)
rgV(y2y1). (14-33)
This work is negative because the upward displacement and the downward gravi-
tational force have opposite directions.
Work must also be done on the system (at the input end) to push the entering
fluid into the tube and by the system (at the output end) to push forward the fluid
that is located ahead of the emerging fluid. In general, the work done by a force
of magnitude F, acting on a fluid sample contained in a tube of area Ato move
the fluid through a distance x,is
Fx(pA)(x)p(Ax)pV.
The work done on the system is then p1V, and the work done by the system
is p2V.Their sum Wpis
Wpp2Vp1V
(p2p1)V. (14-34)
The work–kinetic energy theorem of Eq. 14-31 now becomes
WWgWpK.
Substituting from Eqs. 14-32, 14-33, and 14-34 yields
.
This, after a slight rearrangement, matches Eq. 14-28, which we set out to prove.
rgV(y2y1)V(p2p1)1
2r
V(v2
2v2
1)
(mg
:)
1
2rV(v2
2v2
1)
K1
2mv
2
21
2mv
2
1
Checkpoint 4
Water flows smoothly through the pipe shown in the figure,descending in the process.
Rank the four numbered sections of pipe according to (a) the volume flow rate RV
through them, (b) the flow speed vthrough them, and (c) the water pressure pwithin
them, greatest first.
1
Flow
2
34
Sample Problem 14.06 Bernoulli principle of fluid through a narrowing pipe
Ethanol of density r791 kg/m3flows smoothly through
a horizontal pipe that tapers (as in Fig. 14-15) in cross-
sectional area from A11.20 103m2to A2A1/2.
The pressure difference between the wide and narrow
sections of pipe is 4120 Pa. What is the volume flow rate
RVof the ethanol?
404 CHAPTER 14 FLUIDS
KEY IDEAS
(1) Because the fluid flowing through the wide section of
pipe must entirely pass through the narrow section, the vol-
ume flow rate RVmust be the same in the two sections.Thus,
from Eq. 14-24,
RVv1A1v2A2. (14-35)
However, with two unknown speeds, we cannot evaluate this
equation for RV. (2) Because the flow is smooth, we can ap-
ply Bernoulli’s equation.From Eq.14-28, we can write
, (14-36)
where subscripts 1 and 2 refer to the wide and narrow
sections of pipe, respectively, and yis their common eleva-
tion. This equation hardly seems to help because it does not
contain the desired RVand it contains the unknown speeds
v1and v2.
Calculations: There is a neat way to make Eq. 14-36 work for
us: First, we can use Eq. 14-35 and the fact that A2A1/2 to write
and . (14-37)v2RV
A2
2RV
A1
v1RV
A1
p11
2rv2
1rgy p21
2rv2
2rgy
Sample Problem 14.07 Bernoulli principle for a leaky water tank
In the old West, a desperado fires a bullet into an open water
tank (Fig.14-20), creating a hole a distance hbelow the water
surface.What is the speed vof the water exiting the tank?
KEY IDEAS
(1) This situation is essentially that of water moving (down-
ward) with speed v0through a wide pipe (the tank) of cross-
sectional area Aand then moving (horizontally) with speed v
through a narrow pipe (the hole) of cross-sectional area a. (2)
Because the water flowing through the wide pipe must en-
tirely pass through the narrow pipe, the volume flow rate RV
must be the same in the two “pipes.” (3) We can also relate v
to v0(and to h) through Bernoulli’s equation (Eq. 14-28).
Calculations: From Eq. 14-24,
RVav Av0
and thus
Because aA, we see that v0v.To apply Bernoulli’s equa-
tion, we take the level of the hole as our reference level for
measuring elevations (and thus gravitational potential en-
ergy). Noting that the pressure at the top of the tank and at
the bullet hole is the atmospheric pressure p0(because both
places are exposed to the atmosphere), we write Eq. 14-28 as
(14-39)p01
2rv2
0rgh p01
2rv2rg(0).
v0a
Av.
h
p0y = 0
p
0
Figure 14-20 Water pours
through a hole in a water
tank, at a distance hbelow
the water surface. The pres-
sure at the water surface and
at the hole is atmospheric
pressure p0.
(Here the top of the tank is represented by the left side of
the equation and the hole by the right side.The zero on the
right indicates that the hole is at our reference level.)
Before we solve Eq. 14-39 for v, we can use our result that
v0vto simplify it: We assume that , and thus the term
in Eq. 14-39, is negligible relative to the other terms,
and we drop it. Solving the remaining equation for vthen
yields
(Answer)
This is the same speed that an object would have when
falling a height hfrom rest.
v12gh.
1
2rv2
0
v2
0
Additional examples, video, and practice available at WileyPLUS
Then we can substitute these expressions into Eq. 14-36 to
eliminate the unknown speeds and introduce the desired vol-
ume flow rate.Doing this and solving for RVyield
. (14-38)
We still have a decision to make: We know that the
pressure difference between the two sections is 4120 Pa, but
does that mean that p1p2is 4120 Pa or 4120 Pa? We
could guess the former is true, or otherwise the square root
in Eq. 14-38 would give us an imaginary number. However,
let’s try some reasoning. From Eq. 14-35 we see that speed
v2in the narrow section (small A2) must be greater than
speed v1in the wider section (larger A1). Recall that if the
speed of a fluid increases as the fluid travels along a hori-
zontal path (as here), the pressure of the fluid must
decrease. Thus, p1is greater than p2, and p1p24120 Pa.
Inserting this and known data into Eq. 14-38 gives
2.24 103m3/s. (Answer)
RV1.20 103m2A(2)(4120 Pa)
(3)(791 kg/m3)
RVA1A2( p1p2)
3r
405
QUESTIONS
Density The density rof any material is defined as the material’s
mass per unit volume:
(14-1)
Usually, where a material sample is much larger than atomic
dimensions, we can write Eq. 14-1 as
(14-2)
Fluid Pressure Afluid is a substance that can flow; it conforms
to the boundaries of its container because it cannot withstand shear-
ing stress. It can, however, exert a force perpendicular to its surface.
That force is described in terms of pressure p:
(14-3)
in which Fis the force acting on a surface element of area A. If the
force is uniform over a flat area,Eq. 14-3 can be written as
(14-4)
The force resulting from fluid pressure at a particular point in a
fluid has the same magnitude in all directions. Gauge pressure is the
difference between the actual pressure (or absolute pressure) at a
point and the atmospheric pressure.
Pressure Variation with Height and Depth Pressure in a fluid
at rest varies with vertical position y. For ymeasured positive upward,
p2p1rg(y1y2). (14-7)
The pressure in a fluid is the same for all points at the same level. If
his the depth of a fluid sample below some reference level at which
the pressure is p0, then the pressure in the sample is
pp0rgh. (14-8)
pF
A.
pF
A,
rm
V.
rm
V.
Review & Summary
Pascal’s Principle A change in the pressure applied to an en-
closed fluid is transmitted undiminished to every portion of the
fluid and to the walls of the containing vessel.
Archimedes’ Principle When a body is fully or partially sub-
merged in a fluid, a buoyant force from the surrounding fluid
acts on the body. The force is directed upward and has a magni-
tude given by
Fbmfg, (14-16)
where mfis the mass of the fluid that has been displaced by the body
(that is,the fluid that has been pushed out of the way by the body).
When a body floats in a fluid, the magnitude Fbof the (upward)
buoyant force on the body is equal to the magnitude Fgof the (down-
ward) gravitational force on the body. The apparent weight of a body
on which a buoyant force acts is related to its actual weight by
weightapp weight Fb. (14-19)
Flow of Ideal Fluids An ideal fluid is incompressible and
lacks viscosity, and its flow is steady and irrotational. A streamline
is the path followed by an individual fluid particle.A tube of flow is
a bundle of streamlines.The flow within any tube of flow obeys the
equation of continuity:
RVAv a constant, (14-24)
in which RVis the volume flow rate, Ais the cross-sectional area of
the tube of flow at any point, and vis the speed of the fluid at that
point.The mass flow rate Rmis
RmrRVrAv a constant. (14-25)
Bernoulli’s Equation Applying the principle of conservation
of mechanical energy to the flow of an ideal fluid leads to
Bernoulli’s equation along any tube of flow:
prv2rgy a constant. (14-29)
1
2
F
:
b
1We fully submerge an irregular 3 kg lump of material in a cer-
tain fluid. The fluid that would have been in the space now occu-
pied by the lump has a mass of 2 kg. (a) When we release the lump,
does it move upward, move downward, or remain in place? (b) If
we next fully submerge the lump in a less dense fluid and again re-
lease it, what does it do?
2Figure 14-21 shows four situations in which a red liquid and a gray
liquid are in a U-tube. In one situation the liquids cannot be in static
equilibrium. (a) Which situation is that? (b) For the other three sit-
Questions
(1) (2) (3) (4)
Figure 14-21 Question 2.
uations, assume static equilibrium. For each of them, is the density
of the red liquid greater than, less than, or equal to the density of
the gray liquid?
3A boat with an anchor on board floats in a swimming
pool that is somewhat wider than the boat. Does the pool water
level move up, move down, or remain the same if the anchor is
(a) dropped into the water or (b) thrown onto the surrounding
ground? (c) Does the water level in the pool move upward,
move downward, or remain the
same if, instead, a cork is dropped
from the boat into the water,
where it floats?
4Figure 14-22 shows a tank filled
with water. Five horizontal floors
and ceilings are indicated; all have
the same area and are located at
distances L,2L, or 3Lbelow the
top of the tank. Rank them accord-
ing to the force on them due to the
water,greatest first. Figure 14-22 Question 4.
a
b
e
d
c
water flows smoothly toward the right. The radii of the pipe sec-
tions are indicated. In which arrangements is the net work done on
a unit volume of water moving from the leftmost section to the
rightmost section (a) zero, (b) positive, and (c) negative?
8A rectangular block is pushed
face-down into three liquids, in
turn. The apparent weight Wapp of
the block versus depth hin the
three liquids is plotted in Fig. 14-26.
Rank the liquids according to their
weight per unit volume, greatest
first.
9Water flows smoothly in a hor-
izontal pipe. Figure 14-27 shows
the kinetic energy Kof a water el-
ement as it moves along an xaxis
that runs along the pipe. Rank the
three lettered sections of the pipe
according to the pipe radius, great-
est first.
10 We have three containers with different liquids. The gauge
pressure pgversus depth his plotted in Fig. 14-28 for the liquids.
In each container, we will fully submerge a rigid plastic bead.
Rank the plots according to the magnitude of the buoyant force
on the bead, greatest first.
2.00R2.00RR
(1)
3.00R R 2.00R
(2)
2.00R3.00RR
(
3
)
R R 3.00R
(
4
)
5The teapot effect. Water
poured slowly from a teapot spout
can double back under the spout for
a considerable distance (held there
by atmospheric pressure) before
detaching and falling. In Fig. 14-23,
the four points are at the top or bot-
tom of the water layers, inside or
outside. Rank those four points ac-
cording to the gauge pressure in the water there, most positive first.
6Figure 14-24 shows three identical open-top containers filled to
the brim with water; toy ducks float in two of them. Rank the contain-
ers and contents according to their weight,greatest first.
Figure 14-24 Question 6.
(a) (b) (c)
Figure 14-25 Question 7.
Figure 14-26 Question 8.
W
app
h
a
b
c
406 CHAPTER 14 FLUIDS
W
ater
flow dc
b
a
S
pout
Figure 14-23 Question 5.
K
x
A B C
Figure 14-27 Question 9.
p
g
h
ab
c
Figure 14-28 Question 10.
Module 14-1 Fluids, Density, and Pressure
•1 A fish maintains its depth in fresh water by adjusting the
air content of porous bone or air sacs to make its average density
the same as that of the water. Suppose that with its air sacs col-
lapsed, a fish has a density of 1.08 g/cm3. To what fraction of its ex-
panded body volume must the fish inflate the air sacs to reduce its
density to that of water?
•2 A partially evacuated airtight container has a tight-fitting lid
of surface area 77 m2and negligible mass. If the force required to
remove the lid is 480 N and the atmospheric pressure is 1.0 105
Pa, what is the internal air pressure?
•3 Find the pressure increase in the fluid in a syringe when a
nurse applies a force of 42 N to the syringe’s circular piston, which
has a radius of 1.1 cm.
SSM
ILW
•4 Three liquids that will not mix are poured into a cylindrical con-
tainer.The volumes and densities of the liquids are 0.50 L, 2.6 g/cm3;
0.25 L, 1.0 g/cm3; and 0.40 L, 0.80 g/cm3. What is the force on the
bottom of the container due to these liquids? One liter 1 L
1000 cm3.(Ignore the contribution due to the atmosphere.)
•5 An office window has dimensions 3.4 m by 2.1 m. As a
result of the passage of a storm, the outside air pressure drops to
0.96 atm, but inside the pressure is held at 1.0 atm.What net force
pushes out on the window?
•6 You inflate the front tires on your car to 28 psi.Later,you measure
your blood pressure, obtaining a reading of 120/80, the readings being
in mm Hg.In metric countries (which is to say, most of the world),these
pressures are customarily reported in kilopascals (kPa). In kilopascals,
what are (a) your tire pressure and (b) your blood pressure?
SSM
7Figure 14-25 shows four arrangements of pipes through which
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
407
PROBLEMS
tion against a maximum pressure difference (between inside and
outside the chest cavity) of 0.050 atm.What is the difference in dmax
for fresh water and the water of the Dead Sea (the saltiest natural
water in the world, with a density of 1.5 103kg/m3)?
•13 At a depth of 10.9 km, the Challenger Deep in the
Marianas Trench of the Pacific Ocean is the deepest site in any
ocean. Yet, in 1960, Donald Walsh and Jacques Piccard reached
the Challenger Deep in the bathyscaph Trieste. Assuming that
seawater has a uniform density of 1024 kg/m3, approximate the
hydrostatic pressure (in atmospheres) that the Trieste had to
withstand. (Even a slight defect in the Trieste structure would
have been disastrous.)
•14 Calculate the hydrostatic difference in blood pressure be-
tween the brain and the foot in a person of height 1.83 m.The den-
sity of blood is 1.06 103kg/m3.
•15 What gauge pressure must a machine produce in order to suck
mud of density 1800 kg/m3up a tube by a height of 1.5 m?
•16 Snorkeling by humans
and elephants. When a person
snorkels, the lungs are connected
directly to the atmosphere through
the snorkel tube and thus are at at-
mospheric pressure. In atmo-
spheres, what is the difference
between this internal air pressure
and the water pressure against the
body if the length of the snorkel
tube is (a) 20 cm (standard situation) and (b) 4.0 m (probably
lethal situation)? In the latter, the pressure difference causes
blood vessels on the walls of the lungs to rupture, releasing blood
into the lungs. As depicted in Fig. 14-31, an elephant can safely
snorkel through its trunk while swimming with its lungs 4.0 m be-
low the water surface because the membrane around its lungs
contains connective tissue that holds and protects the blood ves-
sels, preventing rupturing.
•17 Crew members attempt to escape from a dam-
aged submarine 100 m below the surface. What force must be ap-
plied to a pop-out hatch, which is 1.2 m by 0.60 m, to push it out at
that depth? Assume that the density of the ocean water is 1024
kg/m3and the internal air pressure is at 1.00 atm.
•18 In Fig. 14-32, an open tube of length
L1.8 m and cross-sectional area A
4.6 cm2is fixed to the top of a cylindrical bar-
rel of diameter D1.2 m and height H
1.8 m. The barrel and tube are filled with
water (to the top of the tube). Calculate
the ratio of the hydrostatic force on the
bottom of the barrel to the gravitational
force on the water contained in the barrel.
Why is that ratio not equal to 1.0? (You need
not consider the atmospheric pressure.)
••19 A large aquarium of height 5.00
m is filled with fresh water to a depth of
2.00 m. One wall of the aquarium consists
of thick plastic 8.00 m wide. By how much
does the total force on that wall increase
if the aquarium is next filled to a depth of
4.00 m?
SSM
p
••7 In 1654 Otto von Guericke, in-
ventor of the air pump, gave a
demonstration before the noble-
men of the Holy Roman Empire in
which two teams of eight horses
could not pull apart two evacuated
brass hemispheres. (a) Assuming
the hemispheres have (strong) thin
walls, so that Rin Fig. 14-29 may be considered both the inside
and outside radius, show that the force required to pull apart
the hemispheres has magnitude FpR2p, where pis the dif-
ference between the pressures outside and inside the sphere.
(b) Taking Ras 30 cm, the inside pressure as 0.10 atm, and the out-
side pressure as 1.00 atm, find the force magnitude the teams of
horses would have had to exert to pull apart the hemispheres.
(c) Explain why one team of horses could have proved the
point just as well if the hemispheres were attached to a sturdy wall.
Module 14-2 Fluids at Rest
•8 The bends during flight. Anyone who scuba dives is
advised not to fly within the next 24 h because the air mixture
for diving can introduce nitrogen to the bloodstream. Without
allowing the nitrogen to come out of solution slowly, any sudden
air-pressure reduction (such as during airplane ascent) can result
in the nitrogen forming bubbles in the blood, creating the bends,
which can be painful and even fatal. Military special operation
forces are especially at risk. What is the change in pressure on
such a special-op soldier who must scuba dive at a depth of 20 m
in seawater one day and parachute at an altitude of 7.6 km the
next day? Assume that the average air density within the altitude
range is 0.87 kg/m3.
•9 Blood pressure in Argentinosaurus. (a) If this long-
necked, gigantic sauropod had a head height of 21 m and a heart
height of 9.0 m, what (hydrostatic) gauge pressure in its blood
was required at the heart such that the blood pressure at the
brain was 80 torr (just enough to perfuse the brain with blood)?
Assume the blood had a density of . (b) What
was the blood pressure (in torr or mm Hg) at the feet?
•10 The plastic tube in Fig. 14-30 has a
cross-sectional area of 5.00 cm2. The tube is
filled with water until the short arm (of
length d0.800 m) is full.Then the short arm
is sealed and more water is gradually poured
into the long arm. If the seal will pop off when
the force on it exceeds 9.80 N, what total
height of water in the long arm will put the seal
on the verge of popping?
•11 Giraffe bending to drink. In a giraffe with its head 2.0 m
above its heart, and its heart 2.0 m above its feet, the (hydrostatic)
gauge pressure in the blood at its heart is 250 torr.Assume that the gi-
raffe stands upright and the blood density is . In torr
(or mm Hg), find the (gauge) blood pressure (a) at the brain (the
pressure is enough to perfuse the brain with blood,to keep the giraffe
from fainting) and (b) at the feet (the pressure must be countered by
tight-fitting skin acting like a pressure stocking). (c) If the giraffe
were to lower its head to drink from a pond without splaying its legs
and moving slowly, what would be the increase in the blood pressure
in the brain? (Such action would probably be lethal.)
•12 The maximum depth dmax that a diver can snorkel is set
by the density of the water and the fact that human lungs can func-
1.06 103 kg/m3
1.06 103 kg/m3
F
:
R
FF
Figure 14-29 Problem 7.
d
Figure 14-30
Problems 10
and 81.
Figure 14-31 Problem 16.
D
L
H
A
S
U
I
T
A
B
L
E
F
O
R
W
A
T
E
R
F
R
E
S
H
D
R
I
N
K
I
N
G
Figure 14-32
Problem 18.
at depth D35.0 m behind the ver-
tical upstream face of a dam of
width W314 m. Find (a) the net
horizontal force on the dam from
the gauge pressure of the water and
(b) the net torque due to that force
about a horizontal line through O
parallel to the (long) width of the dam.This torque tends to rotate the
dam around that line, which would cause the dam to fail. (c) Find the
moment arm of the torque.
Module 14-3 Measuring Pressure
•25 In one observation, the column in a mercury barometer (as is
shown in Fig. 14-5a) has a measured height hof 740.35 mm.The tem-
perature is 5.0C, at which temperature the density of mercury r is
1.3608 104kg/m3.The free-fall acceleration gat the site of the barom-
408 CHAPTER 14 FLUIDS
eter is 9.7835 m/s2.What is the atmospheric pressure at that site in pas-
cals and in torr (which is the common unit for barometer readings)?
•26 To suck lemonade of density 1000 kg/m3up a straw to a maxi-
mum height of 4.0 cm, what minimum gauge pressure (in atmo-
spheres) must you produce in your lungs?
••27 What would be the height of the atmosphere if the
air density (a) were uniform and (b) decreased linearly to zero
with height? Assume that at sea level the air pressure is 1.0 atm
and the air density is 1.3 kg/m3.
Module 14-4 Pascal’s Principle
•28 A piston of cross-sectional
area ais used in a hydraulic press to
exert a small force of magnitude fon
the enclosed liquid. A connecting
pipe leads to a larger piston of cross-
sectional area A(Fig. 14-36). (a) What
force magnitude Fwill the larger pis-
ton sustain without moving? (b) If
the piston diameters are 3.80 cm and
53.0 cm, what force magnitude on the small piston will balance a 20.0
kN force on the large piston?
••29 In Fig. 14-37, a spring of spring
constant 3.00 104N/m is between a
rigid beam and the output piston of a
hydraulic lever. An empty container
with negligible mass sits on the input
piston.The input piston has area Ai, and
the output piston has area 18.0Ai.
Initially the spring is at its rest length.
How many kilograms of sand must be
(slowly) poured into the container to compress the spring by 5.00 cm?
Module 14-5 Archimedes’ Principle
•30 A 5.00 kg object is released from rest while fully submerged
in a liquid. The liquid displaced by the submerged object has a
mass of 3.00 kg. How far and in what direction does the object
move in 0.200 s, assuming that it moves freely and that the drag
force on it from the liquid is negligible?
•31 A block of wood floats in fresh water with two-thirds of its
volume Vsubmerged and in oil with 0.90Vsubmerged. Find the den-
sity of (a) the wood and (b) the oil.
•32 In Fig. 14-38, a cube of edge
length L0.600 m and mass 450 kg
is suspended by a rope in an open
tank of liquid of density 1030 kg/m3.
Find (a) the magnitude of the total
downward force on the top of the
cube from the liquid and the atmo-
sphere, assuming atmospheric pres-
sure is 1.00 atm, (b) the magnitude
of the total upward force on the bot-
tom of the cube, and (c) the tension
in the rope. (d) Calculate the magnitude of the buoyant force on
the cube using Archimedes’ principle. What relation exists among
all these quantities?
•33 An iron anchor of density 7870 kg/m3appears 200 N
lighter in water than in air. (a) What is the volume of the anchor?
(b) How much does it weigh in air?
•34 A boat floating in fresh water displaces water weighing
SSM
SSM
SSM
••20 The L-shaped fish tank shown in
Fig. 14-33 is filled with water and is open at
the top. If d5.0 m, what is the (total)
force exerted by the water (a) on face A
and (b) on face B?
••21 Two identical cylindrical ves-
sels with their bases at the same level each
contain a liquid of density 1.30 103
kg/m3. The area of each base is 4.00 cm2,
but in one vessel the liquid height is 0.854
m and in the other it is 1.560 m. Find the
work done by the gravitational force in
equalizing the levels when the two vessels are connected.
••22 g-LOC in dogfights.When a pilot takes a tight turn at high
speed in a modern fighter airplane, the blood pressure at the brain
level decreases, blood no longer perfuses the brain, and the blood in
the brain drains.If the heart maintains the (hydrostatic) gauge pressure
in the aorta at 120 torr (or mm Hg) when the pilot undergoes a hori-
zontal centripetal acceleration of 4g, what is the blood pressure (in
torr) at the brain, 30 cm radially inward from the heart? The perfusion
in the brain is small enough that the vision switches to black and white
and narrows to “tunnel vision” and the pilot can undergo g-LOC (“g-
induced loss of consciousness”).Blood density is .
••23 In analyzing certain geo-
logical features, it is often appro-
priate to assume that the pressure
at some horizontal level of com-
pensation, deep inside Earth, is the
same over a large region and is
equal to the pressure due to the
gravitational force on the overly-
ing material. Thus, the pressure on
the level of compensation is given
by the fluid pressure formula. This
model requires, for one thing, that
mountains have roots of continen-
tal rock extending into the denser
mantle (Fig. 14-34). Consider a mountain of height H6.0 km
on a continent of thickness T32 km. The continental rock has
a density of 2.9 g/cm3, and beneath this rock the mantle has a
density of 3.3 g/cm3. Calculate the depth Dof the root. (Hint: Set
the pressure at points aand bequal; the depth yof the level of
compensation will cancel out.)
•••24 In Fig. 14-35, water stands
1.06 103 kg/m3
SSM
Figure 14-36 Problem 28.
aA
F
f
Beam
Spring
Container
Figure 14-37 Problem 29.
L
L/2
Figure 14-38 Problem 32.
d
d
d
d
d
3d
2d
2d
A
B
Figure 14-33
Problem 20.
Mountain
Mantle
3.3 g/cm3
Compensation
level
b a
H
Dy
T
Continent
2.9 g/cm3
Root
Figure 14-34 Problem 23.
O
D
W
Figure 14-35 Problem 24.
409
PROBLEMS
35.6 kN. (a) What is the weight of the water this boat displaces
when floating in salt water of density 1.10 103kg/m3? (b) What is
the difference between the volume of fresh water displaced and
the volume of salt water displaced?
•35 Three children, each of weight 356 N,
make a log raft by lashing together logs of
diameter 0.30 m and length 1.80 m. How
many logs will be needed to keep them
afloat in fresh water? Take the den-
sity of the logs to be 800 kg/m3.
••36 In Fig. 14-39a, a rectan-
gular block is gradually pushed
face-down into a liquid. The block
has height d; on the bottom and
top the face area is A5.67 cm2.
Figure 14-39bgives the apparent
weight Wapp of the block as a func-
tion of the depth hof its lower
face. The scale on the vertical axis
is set by Ws0.20 N. What is the
density of the liquid?
••37 A hollow spherical iron shell floats almost completely sub-
merged in water.The outer diameter is 60.0 cm,and the density of iron
is 7.87 g/cm3. Find the inner diameter.
••38 A small solid ball is
released from rest while fully sub-
merged in a liquid and then its kinetic
energy is measured when it has moved
4.0 cm in the liquid. Figure 14-40 gives
the results after many liquids are used:
The kinetic energy Kis plotted versus
the liquid density rliq, and Ks1.60 J
sets the scale on the vertical axis.
What are (a) the density and (b) the volume of the ball?
••39 A hollow sphere of inner radius 8.0 cm and outer
radius 9.0 cm floats half-submerged in a liquid of density 800 kg/m3.
(a) What is the mass of the sphere? (b) Calculate the density of the
material of which the sphere is made.
••40 Lurking alligators.An al-
ligator waits for prey by floating with
only the top of its head exposed, so
that the prey cannot easily see it.
One way it can adjust the extent of
sinking is by controlling the size of its
lungs. Another way may be by swallowing stones (gastrolithes) that
then reside in the stomach. Figure 14-41 shows a highly simplified
model (a “rhombohedron gater”) of mass 130 kg that roams with its
head partially exposed.The top head surface has area 0.20 m2. If the
alligator were to swallow stones with a total mass of 1.0% of its body
mass (a typical amount), how far would it sink?
••41 What fraction of the volume of an iceberg (density 917 kg/m3)
would be visible if the iceberg floats (a) in the ocean (salt water,den-
sity 1024 kg/m3) and (b) in a river (fresh water, density 1000 kg/m3)?
(When salt water freezes to form ice, the salt is excluded. So, an ice-
berg could provide fresh water to a community.)
••42 A flotation device is in the shape of a right cylinder, with a
height of 0.500 m and a face area of 4.00 m2on top and bottom, and
its density is 0.400 times that of fresh water. It is initially held fully
submerged in fresh water,with its top face at the water surface.Then
WWWSSM
ILW
d
(
a
)
Figure 14-39 Problem 36.
K (J)
K
s
10 2
li
q
(g/cm3)
3
ρ
Figure 14-40 Problem 38.
(b)
Ws
0 1
h (cm)
2
Wapp (N)
Figure 14-41 Problem 40.
it is allowed to ascend gradually until it begins to float. How much
work does the buoyant force do on the device during the ascent?
••43 When researchers find a rea-
sonably complete fossil of a di-
nosaur, they can determine the mass
and weight of the living dinosaur
with a scale model sculpted from
plastic and based on the dimensions
of the fossil bones. The scale of the
model is 1/20; that is, lengths are 1/20
actual length, areas are (1/20)2actual
areas, and volumes are (1/20)3actual
volumes. First, the model is suspended from one arm of a balance
and weights are added to the other arm until equilibrium is
reached. Then the model is fully submerged in water and enough
weights are removed from the second arm to reestablish equilib-
rium (Fig. 14-42). For a model of a particular T. rex fossil, 637.76 g
had to be removed to reestablish equilibrium. What was the vol-
ume of (a) the model and (b) the actual T. rex? (c) If the density of
T. rex was approximately the density of water, what was its mass?
••44 A wood block (mass 3.67 kg, density 600 kg/m3) is fitted
with lead (density 1.14 104kg/m3) so that it floats in water with
0.900 of its volume submerged. Find the lead mass if the lead is fit-
ted to the block’s (a) top and (b) bottom.
••45 An iron casting containing a number of cavities weighs
6000 N in air and 4000 N in water. What is the total cavity volume
in the casting? The density of solid iron is 7.87 g/cm3.
••46 Suppose that you release a small ball from rest at a depth
of 0.600 m below the surface in a pool of water. If the density of the
ball is 0.300 that of water and if the drag force on the ball from the
water is negligible, how high above the water surface will the ball
shoot as it emerges from the water? (Neglect any transfer of en-
ergy to the splashing and waves produced by the emerging ball.)
••47 The volume of air space in the passenger compartment of an
1800 kg car is 5.00 m3. The volume of the motor and front wheels is
0.750 m3,and the volume of the rear wheels, gas tank,and trunk is 0.800
m3; water cannot enter these two regions. The car rolls into a lake. (a)
At first, no water enters the passenger compartment. How much of the
car, in cubic meters, is below the water surface with the car floating
(Fig. 14-43)? (b) As water slowly enters, the car sinks. How many cubic
meters of water are in the car as it disappears below the water surface?
(The car,with a heavy load in the trunk,remains horizontal.)
Figure 14-43 Problem 47.
Figure 14-42 Problem 43.
•••48 Figure 14-44 shows an iron ball suspended by thread of
negligible mass from an upright cylinder that
floats partially submerged in water. The cylin-
der has a height of 6.00 cm, a face area of 12.0
cm2on the top and bottom, and a density of
0.30 g/cm3, and 2.00 cm of its height is above
the water surface.What is the radius of the iron
ball?
Figure 14-44
Problem 48.
with water to a depth D0.30 m. A hole of cross-sectional area
A6.5 cm2in the bottom of the tank allows water to drain out. (a)
What is the drainage rate in cubic meters per second? (b) At what
distance below the bottom of the tank is the cross-sectional area of
the stream equal to one-half the area of the hole?
•58 The intake in Fig. 14-47 has
cross-sectional area of 0.74 m2and
water flow at 0.40 m/s. At the outlet,
distance D180 m below the in-
take, the cross-sectional area is
smaller than at the intake and the
water flows out at 9.5 m/s into
equipment. What is the pressure dif-
ference between inlet and outlet?
•59 Water is moving with a speed of 5.0 m/s through a pipe
with a cross-sectional area of 4.0 cm2.The water gradually descends
10 m as the pipe cross-sectional area increases to 8.0 cm2. (a) What
is the speed at the lower level? (b) If the pressure at the upper level
is 1.5 105Pa, what is the pressure at the lower level?
•60 Models of torpedoes are sometimes tested in a horizontal pipe of
flowing water, much as a wind tunnel is used to test model airplanes.
Consider a circular pipe of internal diameter 25.0 cm and a torpedo
model aligned along the long axis of the pipe.The model has a 5.00 cm
diameter and is to be tested with water flowing past it at 2.50 m/s. (a)
With what speed must the water flow in the part of the pipe that is
unconstricted by the model? (b) What will the pressure difference be
between the constricted and unconstricted parts of the pipe?
•61 A water pipe having a 2.5 cm inside diameter carries wa-
ter into the basement of a house at a speed of 0.90 m/s and a pres-
sure of 170 kPa. If the pipe tapers to 1.2 cm and rises to the second
floor 7.6 m above the input point, what are the (a) speed and
(b) water pressure at the second floor?
••62 A pitot tube (Fig. 14-48) is used to determine the air-
speed of an airplane. It consists of an outer tube with a number of
small holes B(four are shown) that allow air into the tube; that
tube is connected to one arm of a U-tube. The other arm of the
U-tube is connected to hole Aat the front end of the device, which
points in the direction the plane is headed. At Athe air becomes
stagnant so that vA0.At B, however, the speed of the air presum-
ably equals the airspeed vof the plane. (a) Use Bernoulli’s equation
to show that
,
where ris the density of the liquid in the U-tube and his the differ-
ence in the liquid levels in that tube. (b) Suppose that the tube con-
tains alcohol and the level difference his 26.0 cm. What is the
plane’s speed relative to the air? The density of the air is 1.03 kg/m3
and that of alcohol is 810 kg/m3.
vA2rgh
rair
ILW
SSM
410 CHAPTER 14 FLUIDS
Module 14-6 The Equation of Continuity
•49 Canal effect. Figure 14-45
shows an anchored barge that ex-
tends across a canal by distance
and into the water by dis-
tance . The canal has a
width , a water depth
, and a uniform water-flow
speed . Assume that the
flow around the barge is uniform. As
the water passes the bow, the water
level undergoes a dramatic dip
known as the canal effect. If the dip
has depth , what is the water speed alongside the boat
through the vertical cross sections at (a) point aand (b) point b?
The erosion due to the speed increase is a common concern to hy-
draulic engineers.
•50 Figure 14-46 shows two
sections of an old pipe system
that runs through a hill, with
distances dAdB30 m and
D110 m. On each side of
the hill, the pipe radius is
2.00 cm. However, the radius of the pipe inside the hill is no longer
known.To determine it, hydraulic engineers first establish that water
flows through the left and right sections at 2.50 m/s. Then they re-
lease a dye in the water at point Aand find that it takes 88.8 s to
reach point B. What is the average radius of the pipe within the hill?
•51 A garden hose with an internal diameter of 1.9 cm is
connected to a (stationary) lawn sprinkler that consists merely of
a container with 24 holes, each 0.13 cm in diameter. If the water
in the hose has a speed of 0.91 m/s, at what speed does it leave the
sprinkler holes?
•52 Two streams merge to form a river. One stream has a width
of 8.2 m, depth of 3.4 m, and current speed of 2.3 m/s. The other
stream is 6.8 m wide and 3.2 m deep, and flows at 2.6 m/s. If the
river has width 10.5 m and speed 2.9 m/s, what is its depth?
••53 Water is pumped steadily out of a flooded basement at
5.0 m/s through a hose of radius 1.0 cm, passing through a window
3.0 m above the waterline.What is the pump’s power?
••54 The water flowing through a 1.9 cm (inside diameter) pipe
flows out through three 1.3 cm pipes. (a) If the flow rates in the
three smaller pipes are 26, 19, and 11 L/min, what is the flow rate in
the 1.9 cm pipe? (b) What is the ratio of the speed in the 1.9 cm pipe
to that in the pipe carrying 26 L/min?
Module 14-7 Bernoulli’s Equation
•55 How much work is done by pressure in forcing 1.4 m3of
water through a pipe having an internal diameter of 13 mm if the
difference in pressure at the two ends of the pipe is 1.0 atm?
•56 Suppose that two tanks, 1 and 2, each with a large opening at
the top, contain different liquids.A small hole is made in the side of
each tank at the same depth hbelow the liquid surface, but the
hole in tank 1 has half the cross-sectional area of the hole in tank 2.
(a) What is the ratio r1/r2of the densities of the liquids if the mass
flow rate is the same for the two holes? (b) What is the ratio
RV1/RV2of the volume flow rates from the two tanks? (c) At one in-
stant, the liquid in tank 1 is 12.0 cm above the hole. If the tanks are
to have equal volume flow rates, what height above the hole must
the liquid in tank 2 be just then?
SSM
SSM
h0.80 m
vi1.5 m/s
H14 m
D55 m
b12 m
d30 m
•57 A cylindrical tank with a large diameter is filled
SSM
bh
H
Dd
ba i
vi
vi
Figure 14-45 Problem 49.
Figure 14-47 Problem 58.
Reservoir
Generator
building
Outlet
Intake
D
air
ρ
Hole A
Liquid
v
B
ρ
h
B
Air
Figure 14-48 Problems 62 and 63.
dAdB
D
A B
?
Figure 14-46 Problem 50.
411
PROBLEMS
••63 A pitot tube (see Problem 62) on a high-altitude aircraft
measures a differential pressure of 180 Pa. What is the aircraft’s
airspeed if the density of the air is 0.031 kg/m3?
••64 In Fig. 14-49, water flows
through a horizontal pipe and then out
into the atmosphere at a speed v115
m/s. The diameters of the left and right
sections of the pipe are 5.0 cm and 3.0
cm. (a) What volume of water flows
into the atmosphere during a 10 min period? In the left section of the
pipe, what are (b) the speed v2and (c) the gauge pressure?
••65 Aventuri meter is used to measure the flow
speed of a fluid in a pipe. The meter is connected between two
sections of the pipe (Fig. 14-50); the cross-sectional area Aof the
entrance and exit of the meter matches the pipe’s cross-sectional
area. Between the entrance and exit, the fluid flows from the
pipe with speed Vand then through a narrow “throat” of cross-
sectional area awith speed v. A manometer connects the wider
portion of the meter to the narrower portion. The change in the
fluid’s speed is accompanied by a change pin the fluid’s pressure,
which causes a height difference hof the liquid in the two arms of
the manometer.(Here pmeans pressure in the throat minus pres-
sure in the pipe.) (a) By applying Bernoulli’s equation and the
equation of continuity to points 1 and 2 in Fig. 14-50, show that
,
where ris the density of the fluid. (b) Suppose that the fluid is
fresh water, that the cross-sectional areas are 64 cm2in the pipe
and 32 cm2in the throat, and that the pressure is 55 kPa in the pipe
and 41 kPa in the throat. What is the rate of water flow in cubic
meters per second?
VA2a2p
(a2A2)
WWWSSM
opening. (a) Find the magnitude of the
frictional force between plug and pipe
wall. (b) The plug is removed. What
water volume exits the pipe in 3.0 h?
••68 Fresh water flows horizontally
from pipe section 1 of cross-sectional
area A1into pipe section 2 of cross-sec-
tional area A2. Figure 14-52 gives a plot
of the pressure difference p2p1versus
the inverse area squared that
would be expected for a volume flow
rate of a certain value if the water flow
were laminar under all circumstances.
The scale on the vertical axis is set by
Δps300 kN/m2. For the conditions
of the figure, what are the values of
(a) A2and (b) the volume flow rate?
••69 A liquid of density 900 kg/m3
flows through a horizontal pipe that
has a cross-sectional area of 1.90 102m2in region Aand a
A2
1
cross-sectional area of 9.50 102m2in region B. The pressure
difference between the two regions is 7.20 103Pa. What are (a)
the volume flow rate and (b) the mass flow rate?
••70 In Fig. 14-53, water flows
steadily from the left pipe section
(radius r12.00R), through the mid-
dle section (radius R), and into the
right section (radius r33.00R). The
speed of the water in the middle sec-
tion is 0.500 m/s. What is the net work done on 0.400 m3of the wa-
ter as it moves from the left section to the right section?
••71 Figure 14-54 shows a stream of
water flowing through a hole at depth
h10 cm in a tank holding water to
height H40 cm. (a) At what dis-
tance xdoes the stream strike the
floor? (b) At what depth should a sec-
ond hole be made to give the same
value of x? (c) At what depth should a
hole be made to maximize x?
•••72 A very simplified schem-
atic of the rain drainage system for a home is shown in Fig. 14-55.
Rain falling on the slanted roof runs off into gutters around the
roof edge; it then drains through downspouts (only one is
shown) into a main drainage pipe Mbelow the basement, which
d2
v1
v2d1
Figure 14-49 Problem 64.
Aa
12
h
Pipe Pipe
A
M
eter
entrance
M
eter
exit
Venturi meter
Manometer
v
V
Figure 14-50 Problems 65 and 66.
Figure 14-51 Problem 67.
D
d
Figure 14-52 Problem 68.
p2 – p1 (kN/m
2
)
Δ
ps
–Δps
016
A
1
–2 (m–4)
32
r1r3
R
Figure 14-53 Problem 70.
H
h
x
v
Figure 14-54 Problem 71.
••66 Consider the venturi tube of Problem 65 and Fig. 14-50
without the manometer.Let Aequal 5a. Suppose the pressure p1at A
is 2.0 atm. Compute the values of (a) the speed Vat Aand (b) the
speed vat athat make the pressure p2at aequal to zero.(c) Compute
the corresponding volume flow rate if the diameter at Ais 5.0 cm.
The phenomenon that occurs at awhen p2falls to nearly zero is
known as cavitation.The water vaporizes into small bubbles.
••67 In Fig. 14-51, the fresh water behind a reservoir dam
ILW
has depth D15 m. A horizontal pipe 4.0 cm in diameter passes
through the dam at depth d6.0 m. A plug secures the pipe
Floor
drain
w
h1
h2
M
Figure 14-55 Problem 72.
carries the water to an
even larger pipe below the
street. In Fig. 14-55, a floor
drain in the basement is
also connected to drain-
age pipe M. Suppose the
following apply:
(1) the downspouts have
height h111 m, (2) the
floor drain has height h2
1.2 m, (3) pipe Mhas radius
3.0 cm, (4) the house has
side width w30 m and
front length L60 m,(5) all
the water striking the roof goes through pipe M, (6) the initial speed
of the water in a downspout is negligible, and (7) the wind speed is
negligible (the rain falls vertically).
At what rainfall rate, in centimeters per hour, will water from
pipe Mreach the height of the floor drain and threaten to flood
the basement?
Additional Problems
73 About one-third of the body of a person floating in the
Dead Sea will be above the waterline. Assuming that the human
body density is 0.98 g/cm3, find the density of the water in the
Dead Sea. (Why is it so much greater than 1.0 g/cm3?)
74 A simple open U-tube contains mercury. When 11.2 cm of
water is poured into the right arm of the tube, how high above its
initial level does the mercury rise in the left arm?
75 If a bubble in sparkling water accelerates upward at the
rate of 0.225 m/s2and has a radius of 0.500 mm, what is its mass?
Assume that the drag force on the bubble is negligible.
76 Suppose that your body has a uniform density of 0.95
times that of water. (a) If you float in a swimming pool, what frac-
tion of your body’s volume is above the water surface?
Quicksand is a fluid produced when water is forced up into
sand, moving the sand grains away from one another so they are no
longer locked together by friction. Pools of quicksand can form when
water drains underground from hills into valleys where there are
sand pockets. (b) If you float in a deep pool of quicksand that has a
density 1.6 times that of water,what fraction of your body’s volume is
above the quicksand surface? (c) Are you unable to breathe?
77 A glass ball of radius 2.00 cm sits at the bottom of a container
of milk that has a density of 1.03 g/cm3. The normal force on the
ball from the container’s lower surface has magnitude 9.48 102N.
What is the mass of the ball?
78 Caught in an avalanche, a skier is fully submerged in
flowing snow of density 96 kg/m3.Assume that the average density
of the skier, clothing, and skiing equipment is 1020 kg/m3. What
percentage of the gravitational force on the skier is offset by the
buoyant force from the snow?
79 An object hangs from a spring balance.The balance registers
30 N in air, 20 N when this object is immersed in water, and 24 N
when the object is immersed in another liquid of unknown den-
sity.What is the density of that other liquid?
80 In an experiment, a rectangular block with height his allowed
to float in four separate liquids. In the first liquid, which is water, it
floats fully submerged. In liquids A,B, and C, it floats with heights
h/2, 2h/3, and h/4 above the liquid surface, respectively. What are
the relative densities (the densities relative to that of water) of
(a) A, (b) B, and (c) C?
81 Figure 14-30 shows a modified U-tube: the right arm is
shorter than the left arm. The open end of the right arm is height
d10.0 cm above the laboratory bench. The radius throughout
the tube is 1.50 cm.Water is gradually poured into the open end of
the left arm until the water begins to flow out the open end of the
right arm. Then a liquid of density 0.80 g/cm3is gradually added to
the left arm until its height in that arm is 8.0 cm (it does not mix
with the water). How much water flows out of the right arm?
82 What is the acceleration of a rising hot-air balloon if the ratio
of the air density outside the balloon to that inside is 1.39? Neglect
the mass of the balloon fabric and the basket.
SSM
412 CHAPTER 14 FLUIDS
83 Figure 14-56 shows a
siphon, which is a device for
removing liquid from a container.
Tube ABC must initially be filled,
but once this has been done, liquid
will flow through the tube until the
liquid surface in the container is
level with the tube opening at A.
The liquid has density 1000 kg/m3
and negligible viscosity. The dis-
tances shown are h125 cm, d
12 cm, and h240 cm. (a) With
what speed does the liquid emerge
from the tube at C? (b) If the at-
mospheric pressure is 1.0 105Pa,
what is the pressure in the liquid at
the topmost point B? (c) Theoretically, what is the greatest possi-
ble height h1that a siphon can lift water?
84 When you cough, you expel air at high speed through the
trachea and upper bronchi so that the air will remove excess mucus
lining the pathway.You produce the high speed by this procedure:You
breathe in a large amount of air, trap it by closing the glottis (the nar-
row opening in the larynx), increase the air pressure by contracting
the lungs, partially collapse the trachea and upper bronchi to narrow
the pathway, and then expel the air through the pathway by suddenly
reopening the glottis. Assume that during the expulsion the volume
flow rate is 7.0 103m3/s. What multiple of 343 m/s (the speed of
sound vs) is the airspeed through the trachea if the trachea diameter
(a) remains its normal value of 14 mm and (b) contracts to 5.2 mm?
85 A tin can has a total volume of 1200 cm3
and a mass of 130 g. How many grams of lead
shot of density 11.4 g/cm3could it carry with-
out sinking in water?
86 The tension in a string holding a solid
block below the surface of a liquid (of density
greater than the block) is T0when the container
(Fig. 14-57) is at rest. When the container is
given an upward acceleration of 0.250g, what
multiple of T0gives the tension in the string?
87 What is the minimum area (in square meters) of the top sur-
face of an ice slab 0.441 m thick floating on fresh water that will
hold up a 938 kg automobile? Take the densities of ice and fresh
water to be 917 kg/m3and 998 kg/m3, respectively.
88 A 8.60 kg sphere of radius 6.22 cm is at a depth of 2.22 km in
seawater that has an average density of 1025 kg/m3. What are the
(a) gauge pressure, (b) total pressure, and (c) corresponding total
force compressing the sphere’s surface? What are (d) the magni-
tude of the buoyant force on the sphere and (e) the magnitude of
the sphere’s acceleration if it is free to move? Take atmospheric
pressure to be 1.01 105Pa.
89 (a) For seawater of density 1.03 g/cm3, find the weight of wa-
ter on top of a submarine at a depth of 255 m if the horizontal
cross-sectional hull area is 2200.0 m2. (b) In atmospheres, what wa-
ter pressure would a diver experience at this depth?
90 The sewage outlet of a house constructed on a slope is 6.59 m be-
low street level. If the sewer is 2.16 m below street level, find the mini-
mum pressure difference that must be created by the sewage pump to
transfer waste of average density 1000.00 kg/m3from outlet to sewer.
A
B
h2
d
h1
C
Figure 14-56 Problem 83.
Figure 14-57
Problem 86.
413
CHAPTER 15
Oscillations
15-1 SIMPLE HARMONIC MOTION
After reading this module, you should be able to . . .
15.01 Distinguish simple harmonic motion from other types of
periodic motion.
15.02 For a simple harmonic oscillator, apply the relationship
between position xand time tto calculate either if given a
value for the other.
15.03 Relate period T, frequency f, and angular frequency v.
15.04 Identify (displacement) amplitude xm, phase constant
(or phase angle) f, and phase vtf.
15.05 Sketch a graph of the oscillator’s position xversus time
t, identifying amplitude xmand period T.
15.06 From a graph of position versus time, velocity versus
time, or acceleration versus time, determine the amplitude
of the plot and the value of the phase constant f.
15.07 On a graph of position xversus time tdescribe the ef-
fects of changing period T, frequency f, amplitude xm, or
phase constant f.
15.08 Identify the phase constant fthat corresponds to the
starting time (t0) being set when a particle in SHM is
at an extreme point or passing through the center point.
15.09 Given an oscillator’s position x(t)as a function of time,
find its velocity v(t)as a function of time, identify the veloc-
ity amplitude vmin the result, and calculate the velocity at
any given time.
15.10 Sketch a graph of an oscillator’s velocity vversus time t,
identifying the velocity amplitude vm.
15.11 Apply the relationship between velocity amplitude vm,
angular frequency v, and (displacement) amplitude xm.
15.12 Given an oscillator’s velocity v(t)as a function of time,
calculate its acceleration a(t)as a function of time, identify
the acceleration amplitude amin the result, and calculate
the acceleration at any given time.
15.13 Sketch a graph of an oscillator’s acceleration aversus
time t, identifying the acceleration amplitude am.
15.14 Identify that for a simple harmonic oscillator the acceler-
ation aat any instant is always given by the product of a
negative constant and the displacement xjust then.
15.15 For any given instant in an oscillation, apply the relation-
ship between acceleration a, angular frequency v, and dis-
placement x.
15.16 Given data about the position xand velocity vat one
instant, determine the phase vtfand phase constant f.
15.17 For a spring–block oscillator, apply the relationships be-
tween spring constant kand mass mand either period Tor
angular frequency v.
15.18 Apply Hooke’s law to relate the force Fon a simple har-
monic oscillator at any instant to the displacement xof the
oscillator at that instant.
●The frequency fof periodic, or oscillatory, motion is the
number of oscillations per second. In the SI system, it is
measured in hertz: 1Hz1s
1.
●The period Tis the time required for one complete oscilla-
tion, or cycle. It is related to the frequency by T1/f.
●In simple harmonic motion (SHM), the displacement x(t)of a
particle from its equilibrium position is described by the equation
xxmcos(vtf)(displacement),
in which xmis the amplitude of the displacement, vtfis
the phase of the motion, and fis the phase constant. The
angular frequency vis related to the period and frequency of
the motion by v2p/T2pf.
●Differentiating x(t)leads to equations for the particle’s
SHM velocity and acceleration as functions of time:
vvxmsin(vtf)(velocity)
and av2xmcos(vtf)(acceleration).
In the velocity function, the positive quantity vxmis the veloc-
ity amplitude vm. In the acceleration function, the positive
quantity v2xmis the acceleration amplitude am.
●A particle with mass mthat moves under the influence of a
Hooke’s law restoring force given by Fkx is a linear sim-
ple harmonic oscillator with
(angular frequency)
and (period).T2pAm
k
vAk
m
Key Ideas
Learning Objectives
414 CHAPTER 15 OSCILLATIONS
What Is Physics?
Our world is filled with oscillations in which objects move back and forth repeat-
edly. Many oscillations are merely amusing or annoying, but many others are
dangerous or financially important. Here are a few examples: When a bat hits a
baseball, the bat may oscillate enough to sting the batter’s hands or even to break
apart.When wind blows past a power line, the line may oscillate (“gallop” in elec-
trical engineering terms) so severely that it rips apart, shutting off the power
supply to a community. When an airplane is in flight, the turbulence of the air
flowing past the wings makes them oscillate, eventually leading to metal fatigue
and even failure.When a train travels around a curve, its wheels oscillate horizon-
tally (“hunt” in mechanical engineering terms) as they are forced to turn in new
directions (you can hear the oscillations).
When an earthquake occurs near a city, buildings may be set oscillating so
severely that they are shaken apart.When an arrow is shot from a bow, the feathers
at the end of the arrow manage to snake around the bow staff without hitting it be-
cause the arrow oscillates. When a coin drops into a metal collection plate, the coin
oscillates with such a familiar ring that the coin’s denomination can be determined
from the sound. When a rodeo cowboy rides a bull, the cowboy oscillates wildly as
the bull jumps and turns (at least the cowboy hopes to be oscillating).
The study and control of oscillations are two of the primary goals of both
physics and engineering. In this chapter we discuss a basic type of oscillation
called simple harmonic motion.
Heads Up. This material is quite challenging to most students. One reason is
that there is a truckload of definitions and symbols to sort out, but the main reason
is that we need to relate an object’s oscillations (something that we can see or even
experience) to the equations and graphs for the oscillations. Relating the real, visi-
ble motion to the abstraction of an equation or graph requires a lot of hard work.
Simple Harmonic Motion
Figure 15-1 shows a particle that is oscillating about the origin of an xaxis, repeat-
edly going left and right by identical amounts.The frequency fof the oscillation is
the number of times per second that it completes a full oscillation (a cycle) and
has the unit of hertz (abbreviated Hz), where
1 hertz 1 Hz 1 oscillation per second 1s
1. (15-1)
The time for one full cycle is the period Tof the oscillation,which is
.(15-2)
Any motion that repeats at regular intervals is called periodic motion or har-
monic motion. However, here we are interested in a particular type of periodic
motion called simple harmonic motion (SHM). Such motion is a sinusoidal func-
tion of time t. That is, it can be written as a sine or a cosine of time t. Here we
arbitrarily choose the cosine function and write the displacement (or position) of
the particle in Fig.15-1 as
x(t)xmcos(vtf)(displacement), (15-3)
in which xm,v, and fare quantities that we shall define.
Freeze-Frames. Let’s take some freeze-frames of the motion and then arrange
them one after another down the page (Fig. 15-2a). Our first freeze-frame is at t0
when the particle is at its rightmost position on the xaxis. We label that coordi-
nate as xm(the subscript means maximum); it is the symbol in front of the cosine
T1
f
+xm
–xm
x
0
Figure 15-1 A particle repeatedly oscillates
left and right along an xaxis, between
extreme points xmand xm.
415
15-1 SIMPLE HARMONIC MOTION
0+xm
–xm
t= 0
t=T/4
t=T/2
t= 3T/4
t=T
0+xm
–xm
(a)
(c)
x
xm
0
Displacement
Time (t)
(d)
–xm
0+xm
–xm
t= 0
t=T/4
t=T/2
t= 3T/4
t=T
T
0+xm
–xm
(b)
v
v
v
v
0
xm
–xm
0T/2 T
x
xm
0
Displacement
Time (t)
(e)
–xm
A particle oscillates left
and right in simple
harmonic motion.
Rotating the figure reveals
that the motion forms a
cosine function.
This is a graph of the motion,
with the period T indicated.
The speed is zero at
extreme points.
The speed is greatest
at x = 0.
The speed
is zero at the
extreme points.
The speed is greatest
at the midpoint.
Figure 15-2 (a) A sequence of “freeze-frames” (taken at equal time intervals) showing the position of a par-
ticle as it oscillates back and forth about the origin of an xaxis,between the limits xmand xm.(b) The
vector arrows are scaled to indicate the speed of the particle.The speed is maximum when the particle is at
the origin and zero when it is at xm.If the time tis chosen to be zero when the particle is at xm,then the
particle returns to xmat tT,where Tis the period of the motion.The motion is then repeated. (c)
Rotating the figure reveals the motion forms a cosine function of time, as shown in (d). (e) The speed (the
slope) changes.
A
416 CHAPTER 15 OSCILLATIONS
function in Eq. 15-3. In the next freeze-frame, the particle is a bit to the left of xm.
It continues to move in the negative direction of xuntil it reaches the leftmost po-
sition, at coordinate xm. Thereafter, as time takes us down the page through
more freeze-frames, the particle moves back to xmand thereafter repeatedly os-
cillates between xmand xm. In Eq. 15-3, the cosine function itself oscillates be-
tween 1 and l.The value of xmdetermines how far the particle moves in its os-
cillations and is called the amplitude of the oscillations (as labeled in the handy
guide of Fig.15-3).
Figure 15-2bindicates the velocity of the particle with respect to time, in the se-
ries of freeze-frames. We’ll get to a function for the velocity soon, but for now just
notice that the particle comes to a momentary stop at the extreme points and has
its greatest speed (longest velocity vector) as it passes through the center point.
Mentally rotate Fig. 15-2acounterclockwise by 90, so that the freeze-frames
then progress rightward with time. We set time t0 when the particle is at xm.
The particle is back at xmat time tT(the period of the oscillation), when it
starts the next cycle of oscillation. If we filled in lots of the intermediate freeze-
frames and drew a line through the particle positions, we would have the cosine
curve shown in Fig. 15-2d.What we already noted about the speed is displayed in
Fig. 15-2e.What we have in the whole of Fig. 15-2 is a transformation of what we
can see (the reality of an oscillating particle) into the abstraction of a graph. (In
WileyPLUS the transformation of Fig. 15-2 is available as an animation with
voiceover.) Equation 15-3 is a concise way to capture the motion in the abstrac-
tion of an equation.
More Quantities. The handy guide of Fig. 15-3 defines more quantities
about the motion. The argument of the cosine function is called the phase of the
motion. As it varies with time, the value of the cosine function varies. The con-
stant fis called the phase angle or phase constant. It is in the argument only be-
cause we want to use Eq. 15-3 to describe the motion regardless of where the par-
ticle is in its oscillation when we happen to set the clock time to 0. In Fig. 15-2, we set
t0 when the particle is at xm. For that choice, Eq. 15-3 works just fine if we also
set f0. However, if we set t0 when the particle happens to be at some other
location, we need a different value of f. A few values are indicated in Fig. 15-4.
For example, suppose the particle is at its leftmost position when we happen to
start the clock at t0.Then Eq. 15-3 describes the motion if fprad.To check,
substitute t0 and fprad into Eq. 15-3. See, it gives xxmjust then. Now
check the other examples in Fig.15-4.
The quantity vin Eq. 15-3 is the angular frequency of the motion.To relate it
to the frequency fand the period T, let’s first note that the position x(t) of the
particle must (by definition) return to its initial value at the end of a period.That
is, if x(t) is the position at some chosen time t, then the particle must return to that
same position at time tT. Let’s use Eq. 15-3 to express this condition, but let’s
also just set f0 to get it out of the way. Returning to the same position can
then be written as
xmcos vtxmcos v(tT). (15-4)
The cosine function first repeats itself when its argument (the phase, remember)
has increased by 2prad. So, Eq.15-4 tells us that
v(tT)vt2p
or vT2p.
Thus, from Eq. 15-2 the angular frequency is
(15-5)
The SI unit of angular frequency is the radian per second.
v2p
2pf.
Figure 15-3 A handy guide to the quantities
in Eq. 15-3 for simple harmonic motion.
Displacement
at time t
Amplitude
Angular
frequency
Time
Phase
constant
or phase
angle
Phase
x(t) = xm cos( t + )
ω φ
0
+xm
–xm0
prad
prad
3
2
1
2prad
Figure 15-4 Values of fcorresponding to
the position of the particle at time t0.
417
15-1 SIMPLE HARMONIC MOTION
We’ve had a lot of quantities here, quantities that we could experimentally
change to see the effects on the particle’s SHM. Figure 15-5 gives some examples.
The curves in Fig. 15-5ashow the effect of changing the amplitude. Both curves
have the same period. (See how the “peaks” line up?) And both are for f0. (See
how the maxima of the curves both occur at t0?) In Fig. 15-5b, the two curves
have the same amplitude xmbut one has twice the period as the other (and thus half
the frequency as the other). Figure 15-5cis probably more difficult to understand.
The curves have the same amplitude and same period but one is shifted relative to
the other because of the different fvalues. See how the one with f0 is just a reg-
ular cosine curve? The one with the negative fis shifted rightward from it.That is a
general result: negative fvalues shift the regular cosine curve rightward and posi-
tive fvalues shift it leftward. (Try this on a graphing calculator.)
x
x'm
xm
0
Displacement
t
(a)
–xm
–x'm
The amplitudes are different,
but the frequency and
period are the same. x
xm
–xm
0
t
Displacement
T
T' T'
(b)
The amplitudes are the
same, but the frequencies
and periods are different.
–xm
x
xm
0t
Displacement
(c)
φ
= 0
φ
=–_
4
π
This negative value
shifts the cosine
curve rightward.
This zero gives a
regular cosine curve.
Figure 15-5 In all three cases, the blue curve is obtained
from Eq. 15-3 with f0. (a) The red curve differs from
the blue curve only in that the red-curve amplitude x
mis
greater (the red-curve extremes of displacement are high-
er and lower). (b) The red curve differs from the blue
curve only in that the red-curve period is TT/2 (the red
curve is compressed horizontally). (c) The red curve dif-
fers from the blue curve only in that for the red curve
fp/4 rad rather than zero (the negative value of f
shifts the red curve to the right).
Checkpoint 1
A particle undergoing simple harmonic oscillation of period T(like that in Fig. 15-2) is
at xmat time t0. Is it at xm,atxm, at 0, between xmand 0, or between 0 and
xmwhen (a) t2.00T, (b) t3.50T, and (c) t5.25T?
The Velocity of SHM
We briefly discussed velocity as shown in Fig. 15-2b, finding that it varies in magni-
tude and direction as the particle moves between the extreme points (where the
speed is momentarily zero) and through the central point (where the speed is maxi-
mum). To find the velocity v(t) as a function of time, let’s take a time derivative of
the position function x(t) in Eq. 15-3:
or v(t)vxmsin(vtf)(velocity). (15-6)
The velocity depends on time because the sine function varies with time,
between the values of 1 and 1. The quantities in front of the sine function
v(t)dx(t)
dt d
dt [xm cos(vtf)]
418 CHAPTER 15 OSCILLATIONS
determine the extent of the variation in the velocity, between vxmand vxm.
We say that vxmis the velocity amplitude vmof the velocity variation. When the
particle is moving rightward through x0, its velocity is positive and the magni-
tude is at this greatest value. When it is moving leftward through x0, its veloc-
ity is negative and the magnitude is again at this greatest value. This variation
with time (a negative sine function) is displayed in the graph of Fig. 15-6bfor a
phase constant of f0, which corresponds to the cosine function for the dis-
placement versus time shown in Fig.15-6a.
Recall that we use a cosine function for x(t) regardless of the particle’s posi-
tion at t0.We simply choose an appropriate value of fso that Eq. 15-3 gives us
the correct position at t0.That decision about the cosine function leads us to a
negative sine function for the velocity in Eq. 15-6, and the value of fnow gives
the correct velocity at t0.
The Acceleration of SHM
Let’s go one more step by differentiating the velocity function of Eq. 15-6 with
respect to time to get the acceleration function of the particle in simple harmonic
motion:
or a(t)v2xmcos(vtf)(acceleration). (15-7)
We are back to a cosine function but with a minus sign out front. We know the
drill by now.The acceleration varies because the cosine function varies with time,
between 1 and 1. The variation in the magnitude of the acceleration is set by
the acceleration amplitude am, which is the product v2xmthat multiplies the co-
sine function.
Figure 15-6cdisplays Eq. 15-7 for a phase constant f0, consistent with
Figs. 15-6aand 15-6b. Note that the acceleration magnitude is zero when the
cosine is zero, which is when the particle is at x0. And the acceleration mag-
nitude is maximum when the cosine magnitude is maximum, which is when the
particle is at an extreme point, where it has been slowed to a stop so that its
motion can be reversed. Indeed, comparing Eqs. 15-3 and 15-7 we see an extremely
neat relationship:
a(t)v2x(t). (15-8)
This is the hallmark of SHM: (1) The particle’s acceleration is always oppo-
site its displacement (hence the minus sign) and (2) the two quantities are al-
ways related by a constant (v2). If you ever see such a relationship in an oscil-
lating situation (such as with, say, the current in an electrical circuit, or the
rise and fall of water in a tidal bay), you can immediately say that the motion
is SHM and immediately identify the angular frequency vof the motion. In a
nutshell:
a(t)dv(t)
dt d
dt [vxm sin(vtf)]
In SHM, the acceleration ais proportional to the displacement xbut opposite in
sign, and the two quantities are related by the square of the angular frequency v.
Checkpoint 2
Which of the following relationships between a particle’s acceleration aand its
position xindicates simple harmonic oscillation: (a) a3x2, (b) a5x, (c) a4x,
(d) a2/x? For the SHM, what is the angular frequency (assume the unit of rad/s)?
Figure 15-6 (a) The displacement x(t) of a
particle oscillating in SHM with phase
angle fequal to zero. The period Tmarks
one complete oscillation. (b) The velocity
v(t) of the particle. (c) The acceleration
a(t) of the particle.
x
+xm
–xm
0
Displacement
t
T
(a)
v
+ xm
–xm
0
Velocity
t
(b)
ω
ω
a
+2xm
–2xm
0
Acceleration
t
(c)
ω
ω
Extreme
values
here
mean ...
zero
values
here
and ...
extreme
values
here.
419
15-1 SIMPLE HARMONIC MOTION
Figure 15-7 A linear simple harmonic oscil-
lator. The surface is frictionless. Like the
particle of Fig. 15-2, the block moves in
simple harmonic motion once it has been
either pulled or pushed away from the
x0 position and released. Its displace-
ment is then given by Eq. 15-3.
k
x
–xmx= 0 +xm
m
Simple harmonic motion is the motion of a particle when the force acting on it is
proportional to the particle’s displacement but in the opposite direction.
The Force Law for Simple Harmonic Motion
Now that we have an expression for the acceleration in terms of the displacement
in Eq. 15-8, we can apply Newton’s second law to describe the force responsible
for SHM:
Fma m(v2x)(mv2)x. (15-9)
The minus sign means that the direction of the force on the particle is opposite the di-
rection of the displacement of the particle.That is, in SHM the force is a restoring force
in the sense that it fights against the displacement, attempting to restore the particle to
the center point at x0. We’ve seen the general form of Eq. 15-9 back in Chapter 8
when we discussed a block on a spring as in Fig.15-7.There we wrote Hooke’s law,
Fkx, (15-10)
for the force acting on the block. Comparing Eqs. 15-9 and 15-10, we can now re-
late the spring constant k(a measure of the stiffness of the spring) to the mass of
the block and the resulting angular frequency of the SHM:
kmv2. (15-11)
Equation 15-10 is another way to write the hallmark equation for SHM.
Checkpoint 3
Which of the following relationships between the force Fon a particle and the parti-
cle’s position xgives SHM:(a) F5x, (b) F400x2, (c) F10x,(d) F3x2?
The block–spring system of Fig. 15-7 is called a linear simple harmonic oscillator
(linear oscillator, for short), where linear indicates that Fis proportional to xto
the first power (and not to some other power).
If you ever see a situation in which the force in an oscillation is always pro-
portional to the displacement but in the opposite direction, you can immediately
say that the oscillation is SHM. You can also immediately identify the associated
spring constant k. If you know the oscillating mass, you can then determine the
angular frequency of the motion by rewriting Eq.15-11 as
(angular frequency). (15-12)
(This is usually more important than the value of k.) Further, you can determine
the period of the motion by combining Eqs. 15-5 and 15-12 to write
(period). (15-13)
Let’s make a bit of physical sense of Eqs. 15-12 and 15-13. Can you see that a
stiff spring (large k) tends to produce a large v(rapid oscillations) and thus a
small period T? Can you also see that a large mass mtends to result in a small v
(sluggish oscillations) and thus a large period T?
Every oscillating system, be it a diving board or a violin string, has some
element of “springiness” and some element of “inertia” or mass. In Fig. 15-7, these
elements are separated: The springiness is entirely in the spring, which we assume
to be massless, and the inertia is entirely in the block, which we assume to be rigid.
In a violin string,however, the two elements are both within the string.
T2pAm
k
vAk
m
420 CHAPTER 15 OSCILLATIONS
Additional examples, video, and practice available at WileyPLUS
This maximum speed occurs when the oscillating block is
rushing through the origin; compare Figs. 15-6aand 15-6b,
where you can see that the speed is a maximum whenever
x0.
(d) What is the magnitude amof the maximum acceleration
of the block?
KEY IDEA
The magnitude amof the maximum acceleration is the accel-
eration amplitude v2xmin Eq. 15-7.
Calculation: So, we have
amv2xm(9.78 rad/s)2(0.11 m)
11 m/s2. (Answer)
This maximum acceleration occurs when the block is at the
ends of its path, where the block has been slowed to a stop
so that its motion can be reversed. At those extreme
points, the force acting on the block has its maximum mag-
nitude; compare Figs. 15-6aand 15-6c, where you can see
that the magnitudes of the displacement and acceleration
are maximum at the same times, when the speed is zero, as
you can see in Fig. 15-6b.
(e) What is the phase constant ffor the motion?
Calculations: Equation 15-3 gives the displacement of the
block as a function of time. We know that at time t0,
the block is located at xxm. Substituting these initial
conditions, as they are called, into Eq. 15-3 and canceling xm
give us
1cos f. (15-14)
Taking the inverse cosine then yields
f0 rad. (Answer)
(Any angle that is an integer multiple of 2prad also satisfies
Eq. 15-14; we chose the smallest angle.)
(f) What is the displacement function x(t) for the
spring–block system?
Calculation: The function x(t) is given in general form by
Eq. 15-3. Substituting known quantities into that equation
gives us
x(t)xmcos(vtf)
(0.11 m) cos[(9.8 rad/s)t0]
0.11 cos(9.8t), (Answer)
where xis in meters and tis in seconds.
A block whose mass mis 680 g is fastened to a spring whose
spring constant kis 65 N/m. The block is pulled a distance
x11 cm from its equilibrium position at x0 on a fric-
tionless surface and released from rest at t0.
(a) What are the angular frequency, the frequency, and the
period of the resulting motion?
KEY IDEA
The block–spring system forms a linear simple harmonic
oscillator, with the block undergoing SHM.
Calculations: The angular frequency is given by Eq. 15-12:
9.8 rad/s. (Answer)
The frequency follows from Eq.15-5, which yields
(Answer)
The period follows from Eq. 15-2, which yields
(Answer)
(b) What is the amplitude of the oscillation?
KEY IDEA
With no friction involved,the mechanical energy of the spring–
block system is conserved.
Reasoning: The block is released from rest 11 cm from its
equilibrium position, with zero kinetic energy and the
elastic potential energy of the system at a maximum.Thus,
the block will have zero kinetic energy whenever it is
again 11 cm from its equilibrium position, which means it
will never be farther than 11 cm from that position. Its
maximum displacement is 11 cm:
xm11 cm. (Answer)
(c) What is the maximum speed vmof the oscillating block,
and where is the block when it has this speed?
KEY IDEA
The maximum speed vmis the velocity amplitude vxmin Eq. 15-6.
Calculation: Thus, we have
vmvxm(9.78 rad/s)(0.11 m)
1.1 m/s. (Answer)
T1
f1
1.56 Hz 0.64 s 640 ms.
fv
2p9.78 rad/s
2p rad 1.56 Hz 1.6 Hz.
vAk
mA65 N/m
0.68 kg 9.78 rad/s
Sample Problem 15.01 Block–spring SHM, amplitude, acceleration, phase constant
421
15-2 ENERGY IN SIMPLE HARMONIC MOTION
Calculations: We know vand want fand xm. If we divide
Eq. 15-16 by Eq. 15-15, we eliminate one of those unknowns
and reduce the other to a single trig function:
Solving for tan f, we find
0.461.
This equation has two solutions:
f25and f180(25)155.
Normally only the first solution here is displayed by a calcu-
lator, but it may not be the physically possible solution. To
choose the proper solution, we test them both by using them
to compute values for the amplitude xm. From Eq. 15-15, we
find that if f25, then
We find similarly that if f155, then xm0.094 m.
Because the amplitude of SHM must be a positive constant,
the correct phase constant and amplitude here are
f155and xm0.094 m 9.4 cm. (Answer)
xmx(0)
cos f0.0850 m
cos(25)0.094 m.
tan f v(0)
vx(0) 0.920 m/s
(23.5 rad/s)(0.0850 m)
v(0)
x(0) vxmsin f
xmcos fv
tan f.
Sample Problem 15.02 Finding SHM phase constant from displacement and velocity
At t0, the displacement x(0) of the block in a linear oscil-
lator like that of Fig. 15-7 is 8.50 cm. (Read x(0) as “xat
time zero.”) The block’s velocity v(0) then is 0.920 m/s,
and its acceleration a(0) is 47.0 m/s2.
(a) What is the angular frequency vof this system?
KEY IDEA
With the block in SHM, Eqs. 15-3, 15-6, and 15-7 give its dis-
placement, velocity, and acceleration, respectively, and each
contains v.
Calculations: Let’s substitute t0 into each to see
whether we can solve any one of them for v.We find
x(0) xmcos f, (15-15)
v(0) vxmsin f, (15-16)
and a(0) v2xmcos f. (15-17)
In Eq. 15-15, vhas disappeared. In Eqs. 15-16 and 15-17, we
know values for the left sides, but we do not know xmand f.
However, if we divide Eq. 15-17 by Eq. 15-15, we neatly elim-
inate both xmand fand can then solve for vas
23.5 rad/s. (Answer)
(b) What are the phase constant fand amplitude xm?
vAa(0)
x(0) A47.0 m/s2
0.0850 m
Additional examples, video, and practice available at WileyPLUS
15-2 ENERGY IN SIMPLE HARMONIC MOTION
After reading this module, you should be able to . . .
15.19 For a spring–block oscillator, calculate the kinetic energy
and elastic potential energy at any given time.
15.20 Apply the conservation of energy to relate the total en-
ergy of a spring–block oscillator at one instant to the total
energy at another instant.
15.21 Sketch a graph of the kinetic energy, potential energy,
and total energy of a spring–block oscillator, first as a func-
tion of time and then as a function of the oscillator’s position.
15.22 For a spring–block oscillator, determine the block’s po-
sition when the total energy is entirely kinetic energy and
when it is entirely potential energy.
●A particle in simple harmonic motion has, at any time, ki-
netic energy Kmv
2and potential energy Ukx
2. If no
1
2
1
2
friction is present, the mechanical energy EKU
remains constant even though Kand Uchange.
Learning Objectives
Key Ideas
Energy in Simple Harmonic Motion
Let’s now examine the linear oscillator of Chapter 8, where we saw that the energy
transfers back and forth between kinetic energy and potential energy, while the sum
of the two—the mechanical energy Eof the oscillator—remains constant. The
422 CHAPTER 15 OSCILLATIONS
Figure 15-8 (a) Potential energy U(t), kinet-
ic energy K(t), and mechanical energy E
as functions of time tfor a linear harmon-
ic oscillator. Note that all energies are
positive and that the potential energy and
the kinetic energy peak twice during
every period. (b) Potential energy U(x),
kinetic energy K(x), and mechanical energy
Eas functions of position xfor a linear
harmonic oscillator with amplitude xm.
For x0 the energy is all kinetic, and for
xxmit is all potential.
E
nergy
E
0T/2 Tt
K(t)
U(t)
U(t) + K(t)
(a)
x
Energy
K(x)
U(x)
U(x) + K(x)
–xm+xm
0
E
(b)
As time changes, the
energy shifts between
the two types, but the
total is constant.
As position changes, the
energy shifts between
the two types, but the
total is constant.
Checkpoint 4
In Fig. 15-7, the block has a kinetic energy of 3 J and the spring has an elastic potential
energy of 2 J when the block is at x2.0 cm. (a) What is the kinetic energy when
the block is at x0? What is the elastic potential energy when the block is at (b)
x2.0 cm and (c) xxm?
(a) What is the total mechanical energy Eof the spring–
block system?
KEY IDEA
The mechanical energy E(the sum of the kinetic energy
of the block and the potential energy of
the spring) is constant throughout the motion of the oscillator.
Thus, we can evaluate Eat any point during the motion.
Calculations: Because we are given amplitude xmof the os-
cillations, let’s evaluate Ewhen the block is at position xxm,
U1
2kx2
K1
2mv2
Sample Problem 15.03 SHM potential energy, kinetic energy, mass dampers
Many tall buildings have mass dampers, which are anti-sway
devices to prevent them from oscillating in a wind. The de-
vice might be a block oscillating at the end of a spring and
on a lubricated track. If the building sways, say, eastward,
the block also moves eastward but delayed enough so that
when it finally moves, the building is then moving back
westward. Thus, the motion of the oscillator is out of step
with the motion of the building.
Suppose the block has mass m2.72 105kg and is de-
signed to oscillate at frequency f10.0 Hz and with ampli-
tude xm20.0 cm.
potential energy of a linear oscillator like that of Fig. 15-7 is associated entirely
with the spring. Its value depends on how much the spring is stretched or com-
pressed—that is, on x(t).We can use Eqs. 8-11 and 15-3 to find
(15-18)
Caution: A function written in the form cos2A(as here) means (cos A)2and is not
the same as one written cos A2, which means cos(A2).
The kinetic energy of the system of Fig. 15-7 is associated entirely with the
block. Its value depends on how fast the block is moving—that is, on v(t). We can
use Eq. 15-6 to find
(15-19)
If we use Eq. 15-12 to substitute k/mfor v2, we can write Eq. 15-19 as
(15-20)
The mechanical energy follows from Eqs. 15-18 and 15-20 and is
EUK
For any angle a,
cos2asin2a1.
Thus, the quantity in the square brackets above is unity and we have
(15-21)
The mechanical energy of a linear oscillator is indeed constant and independent of
time. The potential energy and kinetic energy of a linear oscillator are shown as
functions of time tin Fig. 15-8aand as functions of displacement xin Fig. 15-8b.In
any oscillating system, an element of springiness is needed to store the potential en-
ergy and an element of inertia is needed to store the kinetic energy.
EUK1
2kx2
m.
1
2kx2
m[cos2(vtf)sin2(vtf)].
1
2kx2
mcos2(vtf)1
2kx2
m sin2(vtf)
K(t)1
2mv21
2kx2
m sin2(vtf).
K(t)1
2mv21
2mv2x2
m sin2(vtf).
U(t)1
2kx21
2kx2
m cos2(vtf).
423
15-3 AN ANGULAR SIMPLE HARMONIC OSCILLATOR
An Angular Simple Harmonic Oscillator
Figure 15-9 shows an angular version of a simple harmonic oscillator; the element
of springiness or elasticity is associated with the twisting of a suspension wire
rather than the extension and compression of a spring as we previously had. The
device is called a torsion pendulum, with torsion referring to the twisting.
If we rotate the disk in Fig. 15-9 by some angular displacement ufrom its rest
position (where the reference line is at u0) and release it, it will oscillate about
that position in angular simple harmonic motion. Rotating the disk through an
angle uin either direction introduces a restoring torque given by
tku. (15-22)
Here k(Greek kappa) is a constant, called the torsion constant, that depends on
the length, diameter, and material of the suspension wire.
Comparison of Eq. 15-22 with Eq. 15-10 leads us to suspect that Eq. 15-22 is
the angular form of Hooke’s law, and that we can transform Eq. 15-13, which
gives the period of linear SHM, into an equation for the period of angular SHM:
We replace the spring constant kin Eq. 15-13 with its equivalent, the constant
Additional examples, video, and practice available at WileyPLUS
15-3 AN ANGULAR SIMPLE HARMONIC OSCILLATOR
After reading this module, you should be able to . . .
15.23 Describe the motion of an angular simple harmonic
oscillator.
15.24 For an angular simple harmonic oscillator, apply the re-
lationship between the torque tand the angular displace-
ment u(from equilibrium).
15.25 For an angular simple harmonic oscillator, apply the re-
lationship between the period T(or frequency f), the rota-
tional inertia I, and the torsion constant k.
15.26 For an angular simple harmonic oscillator at any instant,
apply the relationship between the angular acceleration a, the
angular frequency v, and the angular displacement u.
Learning Objectives
where it has velocity v0. However, to evaluate Uat that
point, we first need to find the spring constant k. From
Eq. 15-12 and Eq.15-5 ,we find
We can now evaluate Eas
(Answer)2.147 107 J 2.1 107 J.
01
2(1.073 109 N/m)(0.20 m)2
EKU1
2mv21
2kx2
1.073 109 N/m.
(2.72 105 kg)(2p)2(10.0 Hz)2
kmv2m(2pf)2
(v2pf)(v2k/m)
(b) What is the block’s speed as it passes through the equi-
librium point?
Calculations: We want the speed at x0, where the
potential energy is and the mechanical energy
is entirely kinetic energy. So, we can write
or v12.6 m/s. (Answer)
Because Eis entirely kinetic energy, this is the maximum
speed vm.
2.147 107 J 1
2(2.72 105 kg)v20,
EKU1
2mv21
2kx2
U1
2kx20
●A torsion pendulum consists of an object suspended on a wire. When the wire is twisted and then released, the object oscil-
lates in angular simple harmonic motion with a period given by
where Iis the rotational inertia of the object about the axis of rotation and kis the torsion constant of the wire.
T2pAI
k ,
Key Idea
Figure 15-9 A torsion pendulum is an angular
version of a linear simple harmonic oscilla-
tor.The disk oscillates in a horizontal plane;
the reference line oscillates with angular
amplitude um.The twist in the suspension
wire stores potential energy as a spring does
and provides the restoring torque.
–m
θ
+m
θ
0
Suspension wire
Fixed end
Reference line
424 CHAPTER 15 OSCILLATIONS
The constant k, which is a property of the wire, is the same for
both figures; only the periods and the rotational inertias differ.
Let us square each of these equations, divide the second
by the first, and solve the resulting equation for Ib.The result is
6.12 104kgm2. (Answer)
IbIa
T2
b
T2
a
(1.73 104 kgm2)(4.76 s)2
(2.53 s)2
T
a2pAIa
k and T
b2pAIb
k.
Sample Problem 15.04 Angular simple harmonic oscillator, rotational inertia, period
Figure 15-10ashows a thin rod whose length Lis 12.4 cm and
whose mass mis 135 g, suspended at its midpoint from a long
wire. Its period Taof angular SHM is measured to be 2.53 s.
An irregularly shaped object, which we call object X, is then
hung from the same wire, as in Fig. 15-10b, and its period Tbis
found to be 4.76 s. What is the rotational inertia of object X
about its suspension axis?
KEY IDEA
The rotational inertia of either the rod or object Xis related
to the measured period by Eq. 15-23.
Calculations: In Table 10-2e, the rotational inertia of a thin
rod about a perpendicular axis through its midpoint is given as
mL2.Thus, we have,for the rod in Fig.15-10a,
IamL2( )(0.135 kg)(0.124 m)2
1.73 104kgm2.
Now let us write Eq. 15-23 twice, once for the rod and once
for object X:
1
12
1
12
1
12
Figure 15-10 Two torsion pen-
dulums, consisting of (a) a
wire and a rod and (b) the
same wire and an irregularly
shaped object.
Suspension
wire
Rod
L
Object X(a) (b)
Additional examples, video, and practice available at WileyPLUS
15-4 PENDULUMS, CIRCULAR MOTION
After reading this module, you should be able to . . .
15.27 Describe the motion of an oscillating simple pendulum.
15.28 Draw a free-body diagram of a pendulum bob with the
pendulum at angle uto the vertical.
15.29 For small-angle oscillations of a simple pendulum, relate
the period T(or frequency f) to the pendulum’s length L.
15.30 Distinguish between a simple pendulum and a physical
pendulum.
15.31 For small-angle oscillations of a physical pendulum, re-
late the period T(or frequency f) to the distance hbe-
tween the pivot and the center of mass.
15.32 For an angular oscillating system, determine the angu-
lar frequency vfrom either an equation relating torque t
and angular displacement uor an equation relating angular
acceleration aand angular displacement u.
15.33 Distinguish between a pendulum’s angular frequency
v(having to do with the rate at which cycles are com-
pleted) and its du/dt (the rate at which its angle with the
vertical changes).
15.34 Given data about the angular position uand rate of
change du/dt at one instant, determine the phase constant f
and amplitude um.
15.35 Describe how the free-fall acceleration can be mea-
sured with a simple pendulum.
15.36 For a given physical pendulum, determine the location
of the center of oscillation and identify the meaning of that
phrase in terms of a simple pendulum.
15.37 Describe how simple harmonic motion is related to uni-
form circular motion.
Learning Objectives
kof Eq. 15-22, and we replace the mass min Eq. 15-13 with its equivalent, the
rotational inertia Iof the oscillating disk.These replacements lead to
(torsion pendulum). (15-23)T2pA
k
425
15-4 PENDULUMS, CIRCULAR MOTION
Pendulums
We turn now to a class of simple harmonic oscillators in which the springiness is
associated with the gravitational force rather than with the elastic properties of
a twisted wire or a compressed or stretched spring.
The Simple Pendulum
If an apple swings on a long thread, does it have simple harmonic motion? If so,
what is the period T? To answer, we consider a simple pendulum, which consists
of a particle of mass m(called the bob of the pendulum) suspended from one end
of an unstretchable, massless string of length Lthat is fixed at the other end, as in
Fig. 15-11a.The bob is free to swing back and forth in the plane of the page, to the
left and right of a vertical line through the pendulum’s pivot point.
The Restoring Torque. The forces acting on the bob are the force from the
string and the gravitational force g, as shown in Fig. 15-11b, where the string makes
an angle with the vertical. We resolve ginto a radial component Fgcos and a
component Fgsin that is tangent to the path taken by the bob.This tangential com-
ponent produces a restoring torque about the pendulum’s pivot point because the
component always acts opposite the displacement of the bob so as to bring the bob
back toward its central location. That location is called the equilibrium position
( ) because the pendulum would be at rest there were it not swinging.
From Eq. 10-41 ,we can write this restoring torque as
tL(Fgsin u), (15-24)
where the minus sign indicates that the torque acts to reduce uand Lis the moment
arm of the force component Fgsin uabout the pivot point. Substituting Eq. 15-24 into
Eq.10-44 (tIa) and then substituting mg as the magnitude of Fg,we obtain
L(mg sin u)Ia, (15-25)
where Iis the pendulum’s rotational inertia about the pivot point and ais its
angular acceleration about that point.
We can simplify Eq. 15-25 if we assume the angle uis small, for then we can
approximate sin uwith u(expressed in radian measure). (As an example, if u
5.000.0873 rad, then sin u0.0872, a difference of only about 0.1%.) With
that approximation and some rearranging,we then have
(15-26)
This equation is the angular equivalent of Eq. 15-8, the hallmark of SHM. It tells
us that the angular acceleration aof the pendulum is proportional to the angular
displacement ubut opposite in sign. Thus, as the pendulum bob moves to the
right, as in Fig. 15-11a, its acceleration to the left increases until the bob stops and
amgL
Iu.
(trF)
u0
u
uF
:
u
F
:T
:
●A simple pendulum consists of a rod of negligible mass that
pivots about its upper end, with a particle (the bob) attached
at its lower end. If the rod swings through only small angles,
its motion is approximately simple harmonic motion with a pe-
riod given by
(simple pendulum),
where Iis the particle’s rotational inertia about the pivot, mis
the particle’s mass, and Lis the rod’s length.
T2pAI
mgL
●A physical pendulum has a more complicated distribution
of mass. For small angles of swinging, its motion is simple
harmonic motion with a period given by
(physical pendulum),
where Iis the pendulum’s rotational inertia about the pivot, m
is the pendulum’s mass, and his the distance between the
pivot and the pendulum’s center of mass.
●Simple harmonic motion corresponds to the projection of
uniform circular motion onto a diameter of the circle.
T2pAI
mgh
Key Ideas
Figure 15-11 (a) A simple pendulum. (b) The
forces acting on the bob are the gravitational
force gand the force from the string.
The tangential component Fgsin uof the
gravitational force is a restoring force that
tends to bring the pendulum back to its cen-
tral position.
T
:
F
:
θ
L
θ
θ
Fg sin
θ
Fg cos
m
s = L
θ
L
m
(a)
(b)
Pivot
point
T
Fg
This
component
merely
pulls on
the string.
This
component
brings the
bob back
to center.
426 CHAPTER 15 OSCILLATIONS
Figure 15-12 A physical pendulum. The
restoring torque is hFgsin u. When u0,
center of mass Changs directly below
pivot point O.
begins moving to the left. Then, when it is to the left of the equilibrium position,
its acceleration to the right tends to return it to the right, and so on, as it swings
back and forth in SHM. More precisely, the motion of a simple pendulum swing-
ing through only small angles is approximately SHM. We can state this
restriction to small angles another way: The angular amplitude umof the motion
(the maximum angle of swing) must be small.
Angular Frequency. Here is a neat trick. Because Eq. 15-26 has the same
form as Eq. 15-8 for SHM, we can immediately identify the pendulum’s angular
frequency as being the square root of the constants in front of the displacement:
.
In the homework problems you might see oscillating systems that do not seem to
resemble pendulums. However, if you can relate the acceleration (linear or angu-
lar) to the displacement (linear or angular), you can then immediately identify
the angular frequency as we have just done here.
Period. Next, if we substitute this expression for vinto Eq. 15-5 ( ),
we see that the period of the pendulum may be written as
(15-27)
All the mass of a simple pendulum is concentrated in the mass mof the particle-
like bob, which is at radius Lfrom the pivot point. Thus, we can use Eq. 10-33
(Imr2) to write ImL2for the rotational inertia of the pendulum.
Substituting this into Eq. 15-27 and simplifying then yield
(simple pendulum, small amplitude). (15-28)
We assume small-angle swinging in this chapter.
The Physical Pendulum
A real pendulum, usually called a physical pendulum, can have a complicated
distribution of mass. Does it also undergo SHM? If so, what is its period?
Figure 15-12 shows an arbitrary physical pendulum displaced to one side
by angle u. The gravitational force gacts at its center of mass C, at a distance h
from the pivot point O. Comparison of Figs. 15-12 and 15-11breveals only one
important difference between an arbitrary physical pendulum and a simple
pendulum. For a physical pendulum the restoring component Fgsin uof the grav-
itational force has a moment arm of distance habout the pivot point, rather than
of string length L. In all other respects, an analysis of the physical pendulum
would duplicate our analysis of the simple pendulum up through Eq. 15-27.
Again (for small um), we would find that the motion is approximately SHM.
If we replace Lwith hin Eq. 15-27, we can write the period as
(physical pendulum, small amplitude). (15-29)
As with the simple pendulum, Iis the rotational inertia of the pendulum about O.
However, now Iis not simply mL2(it depends on the shape of the physical pen-
dulum), but it is still proportional to m.
A physical pendulum will not swing if it pivots at its center of mass.
Formally, this corresponds to putting h0 in Eq. 15-29. That equation then pre-
dicts T:, which implies that such a pendulum will never complete one swing.
T2p
AI
mgh
F
:
T2p
AL
g
T2p
AI
mgL .
v2p/T
vAmgL
I
θ
h
θ
θ θ
Fg sin Fg cos
O
C
Fg
This component
brings the
pendulum
back to center.
427
15-4 PENDULUMS, CIRCULAR MOTION
Corresponding to any physical pendulum that oscillates about a given pivot
point Owith period Tis a simple pendulum of length L0with the same period T.
We can find L0with Eq. 15-28.The point along the physical pendulum at distance
L0from point Ois called the center of oscillation of the physical pendulum for the
given suspension point.
Measuring g
We can use a physical pendulum to measure the free-fall acceleration gat a par-
ticular location on Earth’s surface. (Countless thousands of such measurements
have been made during geophysical prospecting.)
To analyze a simple case, take the pendulum to be a uniform rod of length L,
suspended from one end. For such a pendulum, hin Eq. 15-29, the distance
between the pivot point and the center of mass, is L. Table 10-2etells us that the
rotational inertia of this pendulum about a perpendicular axis through its center
of mass is mL2. From the parallel-axis theorem of Eq. 10-36 (IIcom Mh2),
we then find that the rotational inertia about a perpendicular axis through one
end of the rod is
IIcom mh2mL2m(L)2mL2. (15-30)
If we put hLand ImL2in Eq. 15-29 and solve for g, we find
.(15-31)
Thus, by measuring Land the period T, we can find the value of gat the pendu-
lum’s location. (If precise measurements are to be made, a number of refine-
ments are needed, such as swinging the pendulum in an evacuated chamber.)
g8p2L
3T2
1
3
1
2
1
3
1
2
1
12
1
12
1
2
Checkpoint 5
Three physical pendulums, of masses m0,2m0, and 3m0, have the same shape and size
and are suspended at the same point. Rank the masses according to the periods of the
pendulums, greatest first.
Sample Problem 15.05 Physical pendulum, period and length
In Fig. 15-13a, a meter stick swings about a pivot point at
one end, at distance hfrom the stick’s center of mass.
(a) What is the period of oscillation T?
KEY IDEA
The stick is not a simple pendulum because its mass is not
concentrated in a bob at the end opposite the pivot point—
so the stick is a physical pendulum.
Calculations: The period for a physical pendulum is
given by Eq. 15-29, for which we need the rotational
inertia Iof the stick about the pivot point. We can treat
the stick as a uniform rod of length Land mass m. Then
Eq. 15-30 tells us that ImL2, and the distance hin
1
3
Figure 15-13 (a) A meter stick suspended from one end as a
physical pendulum. (b) A simple pendulum whose length L0is
chosen so that the periods of the two pendulums are equal.
Point Pon the pendulum of (a) marks the center of oscillation.
P
C
h
L0
(a) (b)
O
Eq. 15-29 is L. Substituting these quantities into Eq. 15-29,
1
2
428 CHAPTER 15 OSCILLATIONS
we find
(15-32)
(15-33)
(Answer)
Note the result is independent of the pendulum’s mass m.
(b) What is the distance L0between the pivot point Oof the
stick and the center of oscillation of the stick?
Calculations: We want the length L0of the simple pendu-
2pA(2)(1.00 m)
(3)(9.8 m/s2)1.64 s.
2p
A2L
3g
T2p
AI
mgh 2pA1
3mL2
mg(1
2L)
Additional examples, video, and practice available at WileyPLUS
Simple Harmonic Motion and Uniform Circular Motion
In 1610, Galileo, using his newly constructed telescope, discovered the four prin-
cipal moons of Jupiter. Over weeks of observation, each moon seemed to him to
be moving back and forth relative to the planet in what today we would call
simple harmonic motion; the disk of the planet was the midpoint of the motion.
The record of Galileo’s observations, written in his own hand, is actually still
available.A. P. French of MIT used Galileo’s data to work out the position of the
moon Callisto relative to Jupiter (actually, the angular distance from Jupiter as
seen from Earth) and found that the data approximates the curve shown in Fig.
15-14. The curve strongly suggests Eq. 15-3, the displacement function for simple
harmonic motion.A period of about 16.8 days can be measured from the plot, but
it is a period of what exactly? After all, a moon cannot possibly be oscillating back
and forth like a block on the end of a spring, and so why would Eq. 15-3 have
anything to do with it?
Actually, Callisto moves with essentially constant speed in an essentially cir-
cular orbit around Jupiter. Its true motion—far from being simple harmonic—
is uniform circular motion along that orbit.What Galileo saw—and what you can
see with a good pair of binoculars and a little patience—is the projection of this
uniform circular motion on a line in the plane of the motion. We are led by
Galileo’s remarkable observations to the conclusion that simple harmonic
lum (drawn in Fig. 15-13b) that has the same period as the
physical pendulum (the stick) of Fig. 15-13a. Setting Eqs.
15-28 and 15-33 equal yields
(15-34)
You can see by inspection that
L0L(15-35)
( )(100 cm) 66.7 cm. (Answer)
In Fig. 15-13a, point Pmarks this distance from suspension
point O. Thus, point Pis the stick’s center of oscillation for
the given suspension point. Point Pwould be different for a
different suspension choice.
2
3
2
3
2p
A2L
3g.T2pAL0
g
Figure 15-14 The angle between Jupiter and its moon Callisto as seen from Earth. Galileo’s 1610
measurements approximate this curve, which suggests simple harmonic motion. At Jupiter’s
mean distance from Earth, 10 minutes of arc corresponds to about 2 106km. (Based on A. P.
French, Newtonian Mechanics, W. W. Norton & Company, New York, 1971, p. 288.)
t(days)
Angle (arc minutes)
12
0010 20 30
⫺12
40
429
15-4 PENDULUMS, CIRCULAR MOTION
motion is uniform circular motion viewed edge-on. In more formal language:
ω
ω
ω
O
y
x
φ
P'
P
v(t)
φ
xm
t+
t+
v
This relates the
velocities of
Pand P´.
O
y
x
P'
P
a(t)
2xm
t+
φω
ω
a
This relates the
accelerations of
Pand P´.
O
y
x
t +
φω
x
m
P'
P
x(t)
P´ is a particle
moving in a circle.
P is a projection
moving in SHM.
(b)(c)
(a)
Figure 15-15 (a) A reference particle Pmoving with uniform circular motion in a reference
circle of radius xm. Its projection Pon the xaxis executes simple harmonic motion. (b) The
projection of the velocity of the reference particle is the velocity of SHM. (c) The projec-
tion of the radial acceleration of the reference particle is the acceleration of SHM.
a
:
v
:
Simple harmonic motion is the projection of uniform circular motion on a diame-
ter of the circle in which the circular motion occurs.
Figure 15-15agives an example. It shows a reference particle Pmoving in
uniform circular motion with (constant) angular speed vin a reference circle. The
radius xmof the circle is the magnitude of the particle’s position vector. At any
time t, the angular position of the particle is vtf, where fis its angular posi-
tion at t0.
Position. The projection of particle Ponto the xaxis is a point P, which we
take to be a second particle. The projection of the position vector of particle P
onto the xaxis gives the location x(t) of P. (Can you see the xcomponent in the
triangle in Fig.15-5a?) Thus, we find
x(t)xmcos(vtf), (15-36)
which is precisely Eq. 15-3. Our conclusion is correct. If reference particle P
moves in uniform circular motion, its projection particle Pmoves in simple
harmonic motion along a diameter of the circle.
Velocity. Figure 15-15bshows the velocity of the reference particle. From
Eq. 10-18 (vr), the magnitude of the velocity vector is xm; its projection on
the xaxis is
v(t)vxmsin(vtf), (15-37)
which is exactly Eq. 15-6.The minus sign appears because the velocity component
of Pin Fig. 15-15bis directed to the left, in the negative direction of x. (The minus
sign is consistent with the derivative of Eq. 15-36 with respect to time.)
Acceleration. Figure 15-15cshows the radial acceleration of the reference
particle. From Eq. 10-23 (arr), the magnitude of the radial acceleration vec-
tor is 2xm; its projection on the xaxis is
a(t)v2xmcos(vtf), (15-38)
which is exactly Eq. 15-7.Thus, whether we look at the displacement, the velocity,
or the acceleration, the projection of uniform circular motion is indeed simple
harmonic motion.
v
v2
a
:
vv
v
:
430 CHAPTER 15 OSCILLATIONS
15-5 DAMPED SIMPLE HARMONIC MOTION
After reading this module, you should be able to . . .
15.38 Describe the motion of a damped simple harmonic oscil-
lator and sketch a graph of the oscillator’s position as a func-
tion of time.
15.39 For any particular time, calculate the position of a
damped simple harmonic oscillator.
15.40 Determine the amplitude of a damped simple harmonic
oscillator at any given time.
15.41 Calculate the angular frequency of a damped simple
harmonic oscillator in terms of the spring constant, the
damping constant, and the mass, and approximate the
angular frequency when the damping constant is small.
15.42 Apply the equation giving the (approximate) total
energy of a damped simple harmonic oscillator as a func-
tion of time.
●The mechanical energy Ein a real oscillating system de-
creases during the oscillations because external forces, such
as a drag force, inhibit the oscillations and transfer mechani-
cal energy to thermal energy. The real oscillator and its motion
are then said to be damped.
●If the damping force is given by db, where is the
velocity of the oscillator and bis a damping constant, then the
displacement of the oscillator is given by
x(t)xmebt/2mcos(vtf),
v
:
v
:
F
:
where v, the angular frequency of the damped oscillator, is
given by
●If the damping constant is small (b), then vv,
where vis the angular frequency of the undamped oscillator.
For small b, the mechanical energy Eof the oscillator is given by
E(t)kx2
mebt/m.
1
2
1km
Ak
mb2
4m2.
Learning Objectives
Key Ideas
Figure 15-16 An idealized damped simple
harmonic oscillator.A vane immersed in a
liquid exerts a damping force on the block
as the block oscillates parallel to the xaxis.
Springiness, k
Rigid support
Mass m
Damping, b
Vane
x
Damped Simple Harmonic Motion
A pendulum will swing only briefly underwater, because the water exerts on the
pendulum a drag force that quickly eliminates the motion.A pendulum swinging
in air does better, but still the motion dies out eventually, because the air exerts
a drag force on the pendulum (and friction acts at its support point), transferring
energy from the pendulum’s motion.
When the motion of an oscillator is reduced by an external force, the oscil-
lator and its motion are said to be damped. An idealized example of a damped
oscillator is shown in Fig. 15-16, where a block with mass moscillates vertically on
a spring with spring constant k. From the block, a rod extends to a vane (both
assumed massless) that is submerged in a liquid.As the vane moves up and down,
the liquid exerts an inhibiting drag force on it and thus on the entire oscillating
system. With time, the mechanical energy of the block–spring system decreases,
as energy is transferred to thermal energy of the liquid and vane.
Let us assume the liquid exerts a damping force that is proportional to the
velocity of the vane and block (an assumption that is accurate if the vane
moves slowly). Then, for force and velocity components along the xaxis in Fig.
15-16, we have
Fdbv, (15-39)
where bis a damping constant that depends on the characteristics of both the
vane and the liquid and has the SI unit of kilogram per second. The minus sign
indicates that opposes the motion.F
:
d
v
:
F
:
d
Damped Oscillations. The force on the block from the spring is Fskx.
Let us assume that the gravitational force on the block is negligible relative to Fd
and Fs. Then we can write Newton’s second law for components along the xaxis
(Fnet,xmax) as
bv kx ma. (15-40)
431
15-5 DAMPED SIMPLE HARMONIC MOTION
Substituting dx/dt for vand d2x/dt2for aand rearranging give us the differential
equation
.(15-41)
The solution of this equation is
x(t)xmebt/2mcos(vtf), (15-42)
where xmis the amplitude and vis the angular frequency of the damped oscilla-
tor.This angular frequency is given by
(15-43)
If b0 (there is no damping), then Eq. 15-43 reduces to Eq. 15-12 ( )
for the angular frequency of an undamped oscillator, and Eq. 15-42 reduces to
Eq. 15-3 for the displacement of an undamped oscillator. If the damping constant
is small but not zero (so that b), then vv.
Damped Energy. We can regard Eq. 15-42 as a cosine function whose amplitude,
which is xmebt/2m, gradually decreases with time, as Fig. 15-17 suggests. For an un-
damped oscillator, the mechanical energy is constant and is given by Eq. 15-21 (E
kx2
m).If the oscillator is damped, the mechanical energy is not constant but decreases
with time. If the damping is small, we can find E(t) by replacing xmin Eq. 15-21 with
xmebt/2m,the amplitude of the damped oscillations.By doing so,we find that
E(t)kx2
mebt/m, (15-44)
which tells us that, like the amplitude, the mechanical energy decreases exponen-
tially with time.
1
2
1
2
1km
v1k/m
vAk
mb2
4m2.
md2x
dt2bdx
dt kx 0
Figure 15-17 The displacement function x(t) for the damped oscillator of Fig. 15-16. The ampli-
tude, which is xmebt/2m, decreases exponentially with time.
t (s)
01 2 3 4 5 6
+xm
–xmxme–bt/2m
xme–bt/2m
x(t)
x
Checkpoint 6
Here are three sets of values for the spring constant, damping constant, and mass
for the damped oscillator of Fig. 15-16. Rank the sets according to the time re-
quired for the mechanical energy to decrease to one-fourth of its initial value,
greatest first.
Set 1 2k0b0m0
Set 2 k06b04m0
Set 3 3k03b0m0
432 CHAPTER 15 OSCILLATIONS
on the left side.Thus,
5.0 s. (Answer)
Because T0.34 s, this is about 15 periods of oscillation.
(c) How long does it take for the mechanical energy to drop
to one-half its initial value?
KEY IDEA
From Eq. 15-44, the mechanical energy at time tis kx2
mebt/m.
Calculations: The mechanical energy has the value
kx2
mat t0.Thus, we must find the value of tfor which
.
If we divide both sides of this equation by kx2
mand solve for
tas we did above, we find
(Answer)
This is exactly half the time we calculated in (b), or about
7.5 periods of oscillation. Figure 15-17 was drawn to illus-
trate this sample problem.
tm ln 1
2
b(0.25 kg)(ln 1
2)
0.070 kg/s 2.5 s.
1
2
1
2kx2
mebt/m1
2(1
2kx2
m)
1
2
1
2
t2m ln 1
2
b(2)(0.25 kg)(ln 1
2)
0.070 kg/s
Sample Problem 15.06 Damped harmonic oscillator, time to decay, energy
For the damped oscillator of Fig. 15-16, m250 g, k
85 N/m, and b70 g/s.
(a) What is the period of the motion?
KEY IDEA
Because b4.6 kg/s, the period is approximately
that of the undamped oscillator.
Calculation: From Eq.15-13, we then have
(Answer)
(b) How long does it take for the amplitude of the damped
oscillations to drop to half its initial value?
KEY IDEA
The amplitude at time tis displayed in Eq. 15-42 as xmebt/2m.
Calculations: The amplitude has the value xmat t0.Thus,
we must find the value of tfor which
xmebt/2mxm.
Canceling xmand taking the natural logarithm of the equa-
tion that remains, we have ln on the right side and
ln(ebt/2m)bt/2m
1
2
1
2
T2p
Am
k2p
A0.25 kg
85 N/m 0.34 s.
1km
Additional examples, video, and practice available at WileyPLUS
15-6 FORCED OSCILLATIONS AND RESONANCE
After reading this module, you should be able to . . .
15.43 Distinguish between natural angular frequency vand
driving angular frequency vd.
15.44 For a forced oscillator, sketch a graph of the oscillation
amplitude versus the ratio vd/vof driving angular fre-
quency to natural angular frequency, identify the approxi-
mate location of resonance, and indicate the effect of in-
creasing the damping constant.
15.45 For a given natural angular frequency v, identify the ap-
proximate driving angular frequency vdthat gives resonance.
●If an external driving force with angular frequency vdacts
on an oscillating system with natural angular frequency v, the
system oscillates with angular frequency vd.
●The velocity amplitude vmof the system is greatest when
vdv,
a condition called resonance. The amplitude xmof the system
is (approximately) greatest under the same condition.
Learning Objectives
Key Ideas
Forced Oscillations and Resonance
A person swinging in a swing without anyone pushing it is an example of free
oscillation. However, if someone pushes the swing periodically, the swing has
433
15-6 FORCED OSCILLATIONS AND RESONANCE
forced, or driven, oscillations. Two angular frequencies are associated with a sys-
tem undergoing driven oscillations: (1) the natural angular frequency vof the
system, which is the angular frequency at which it would oscillate if it were
suddenly disturbed and then left to oscillate freely, and (2) the angular frequency
vdof the external driving force causing the driven oscillations.
We can use Fig. 15-16 to represent an idealized forced simple harmonic oscil-
lator if we allow the structure marked “rigid support” to move up and down at
a variable angular frequency vd. Such a forced oscillator oscillates at the angular
frequency vdof the driving force,and its displacement x(t) is given by
x(t)xmcos(vdtf), (15-45)
where xmis the amplitude of the oscillations.
How large the displacement amplitude xmis depends on a complicated
function of vdand v. The velocity amplitude vmof the oscillations is easier to
describe: it is greatest when
vdv(resonance), (15-46)
a condition called resonance. Equation 15-46 is also approximately the condition
at which the displacement amplitude xmof the oscillations is greatest.Thus, if you
push a swing at its natural angular frequency, the displacement and velocity
amplitudes will increase to large values, a fact that children learn quickly by trial
and error. If you push at other angular frequencies, either higher or lower, the
displacement and velocity amplitudes will be smaller.
Figure 15-18 shows how the displacement amplitude of an oscillator de-
pends on the angular frequency vdof the driving force, for three values of
the damping coefficient b. Note that for all three the amplitude is approxi-
mately greatest when vd/v1 (the resonance condition of Eq. 15-46). The
curves of Fig. 15-18 show that less damping gives a taller and narrower reso-
nance peak.
Examples. All mechanical structures have one or more natural angular fre-
quencies, and if a structure is subjected to a strong external driving force that
matches one of these angular frequencies, the resulting oscillations of the struc-
ture may rupture it. Thus, for example, aircraft designers must make sure that
none of the natural angular frequencies at which a wing can oscillate matches the
angular frequency of the engines in flight. A wing that flaps violently at certain
engine speeds would obviously be dangerous.
Resonance appears to be one reason buildings in Mexico City collapsed in
September 1985 when a major earthquake (8.1 on the Richter scale) occurred
on the western coast of Mexico.The seismic waves from the earthquake should
have been too weak to cause extensive damage when they reached Mexico
City about 400 km away. However, Mexico City is largely built on an ancient
lake bed, where the soil is still soft with water. Although the amplitude of the
seismic waves was small in the firmer ground en route to Mexico City, their
amplitude substantially increased in the loose soil of the city. Acceleration am-
plitudes of the waves were as much as 0.20g, and the angular frequency was
(surprisingly) concentrated around 3 rad/s. Not only was the ground severely
oscillated, but many intermediate-height buildings had resonant angular fre-
quencies of about 3 rad/s. Most of those buildings collapsed during the shaking
(Fig. 15-19), while shorter buildings (with higher resonant angular frequen-
cies) and taller buildings (with lower resonant angular frequencies) remained
standing.
During a 1989 earthquake in the San Francisco–Oakland area, a similar
resonant oscillation collapsed part of a freeway, dropping an upper deck
onto a lower deck. That section of the freeway had been constructed on a
loosely structured mudfill.
Figure 15-18 The displacement amplitude xm
of a forced oscillator varies as the angular
frequency vdof the driving force is varied.
The curves here correspond to three val-
ues of the damping constant b.
Amplitude
0.6 0.8 1.0 1.2 1.4
ω
d
/
ω
b = 50 g/s
(least
damping)
b = 70 g/s
b= 140 g/s
Figure 15-19 In 1985, buildings of intermedi-
ate height collapsed in Mexico City as a
result of an earthquake far from the city.
Taller and shorter buildings remained
standing.
John T. Barr/Getty Images, Inc.
434 CHAPTER 15 OSCILLATIONS
Frequency The frequency f of periodic, or oscillatory, motion is
the number of oscillations per second. In the SI system, it is mea-
sured in hertz:
1 hertz 1Hz1 oscillation per second 1s
1. (15-1)
Period The period T is the time required for one complete oscil-
lation, or cycle. It is related to the frequency by
(15-2)
Simple Harmonic Motion In simple harmonic motion (SHM),
the displacement x(t) of a particle from its equilibrium position is
described by the equation
xxmcos(vtf)(displacement), (15-3)
in which xmis the amplitude of the displacement, vtfis the
phase of the motion, and fis the phase constant. The angular fre-
quency vis related to the period and frequency of the motion by
(angular frequency). (15-5)
Differentiating Eq. 15-3 leads to equations for the particle’s SHM
velocity and acceleration as functions of time:
vvxmsin(vtf)(velocity) (15-6)
and av2xmcos(vtf)(acceleration). (15-7)
In Eq. 15-6, the positive quantity vxmis the velocity amplitude vm
of the motion. In Eq. 15-7, the positive quantity v2xmis the acceler-
ation amplitude amof the motion.
The Linear Oscillator A particle with mass mthat moves un-
der the influence of a Hooke’s law restoring force given by F
kx exhibits simple harmonic motion with
(angular frequency) (15-12)
and (period). (15-13)
Such a system is called a linear simple harmonic oscillator.
Energy A particle in simple harmonic motion has, at any time,
kinetic energy Kmv
2and potential energy Ukx
2. If no fric-
tion is present, the mechanical energy EKUremains con-
stant even though Kand Uchange.
1
2
1
2
T2pAm
k
vAk
m
v2p
T2pf
T1
f.
Review & Summary
Pendulums Examples of devices that undergo simple
harmonic motion are the torsion pendulum of Fig. 15-9, the simple
pendulum of Fig. 15-11, and the physical pendulum of Fig. 15-12.
Their periods of oscillation for small oscillations are, respectively,
(torsion pendulum), (15-23)
(simple pendulum), (15-28)
(physical pendulum). (15-29)
Simple Harmonic Motion and Uniform Circular Motion
Simple harmonic motion is the projection of uniform circular
motion onto the diameter of the circle in which the circular motion
occurs. Figure 15-15 shows that all parameters of circular motion
(position, velocity, and acceleration) project to the corresponding
values for simple harmonic motion.
Damped Harmonic Motion The mechanical energy Ein a real
oscillating system decreases during the oscillations because external
forces, such as a drag force, inhibit the oscillations and transfer me-
chanical energy to thermal energy. The real oscillator and its motion
are then said to be damped. If the damping force is given by d
b, where is the velocity of the oscillator and bis a damping con-
stant, then the displacement of the oscillator is given by
x(t)xmebt/2mcos(vtf), (15-42)
where v, the angular frequency of the damped oscillator, is
given by
(15-43)
If the damping constant is small (b), then vv, where v
is the angular frequency of the undamped oscillator. For small b,
the mechanical energy Eof the oscillator is given by
E(t)kx2
mebt/m. (15-44)
Forced Oscillations and Resonance If an external
driving force with angular frequency vdacts on an oscillating sys-
tem with natural angular frequency v, the system oscillates with
angular frequency vd. The velocity amplitude vmof the system is
greatest when
vdv, (15-46)
a condition called resonance. The amplitude xmof the system is
(approximately) greatest under the same condition.
1
2
1km
vAk
mb2
4m2.
v
:
v
:
F
:
T2p
2I/mgh
T2p
2L/g
T2p
2I/k
Questions
1Which of the following describe ffor the SHM of Fig. 15-20a:
(a) pfp/2,
(b) pf3p/2,
(c) 3p/2 fp?
2The velocity v(t) of a particle undergoing SHM is graphed in
Fig. 15-20b. Is the particle momentarily stationary, headed toward
xm, or headed toward xmat (a) point Aon the graph and (b)
point B? Is the particle at xm, at xm, at 0, between xmand 0, or
between 0 and xmwhen its velocity is represented by (c) point A
and (d) point B? Is the speed of the particle increasing or decreas-
ing at (e) point Aand (f) point B?
x
t
v
t
A
B
(a) (b)
Figure 15-20 Questions 1 and 2.
435
11 In Fig. 15-28, a spring–block
system is put into SHM in two ex-
periments. In the first, the block is
pulled from the equilibrium position
through a displacement d1and then
released. In the second, it is pulled
from the equilibrium position
through a greater displacement d2and then released. Are the
(a) amplitude, (b) period, (c) frequency, (d) maximum kinetic en-
ergy, and (e) maximum potential energy in the second experiment
greater than, less than, or the same as those in the first experiment?
12 Figure 15-29 gives, for three situations, the displacements x(t)
of a pair of simple harmonic oscillators (Aand B) that are identical
except for phase. For each pair, what phase shift (in radians and in
degrees) is needed to shift the curve for Ato coincide with the
curve for B? Of the many possible answers, choose the shift with
the smallest absolute magnitude.
QUESTIONS
3The acceleration a(t) of a parti-
cle undergoing SHM is graphed in
Fig. 15-21. (a) Which of the labeled
points corresponds to the particle
at xm? (b) At point 4, is the veloc-
ity of the particle positive, negative,
or zero? (c) At point 5, is the parti-
cle at xm, at xm, at 0, between
xmand 0, or between 0 and xm?
4Which of the following relationships between the acceleration
aand the displacement xof a particle involve SHM: (a) a0.5x,
(b) a400x2, (c) a20x, (d) a3x2?
5You are to complete Fig. 15-22a
so that it is a plot of velocity vversus
time tfor the spring–block oscillator
that is shown in Fig. 15-22bfor t0.
(a) In Fig. 15-22a, at which lettered
point or in what region between the
points should the (vertical) vaxis in-
tersect the taxis? (For example,
should it intersect at point A,or
maybe in the region between points
Aand B?) (b) If the block’s veloc-
ity is given by vvmsin(vtf),
what is the value of f? Make it pos-
itive, and if you cannot specify the
value (such as p/2 rad), then give
a range of values (such as between
0 and p/2 rad).
6You are to complete Fig. 15-23a
so that it is a plot of acceleration a
versus time tfor the spring–block
oscillator that is shown in Fig. 15-
23bfor t0. (a) In Fig. 15-23a,at
which lettered point or in what re-
gion between the points should the
(vertical) aaxis intersect the taxis?
(For example, should it intersect at
point A, or maybe in the region be-
tween points Aand B?) (b) If the
block’s acceleration is given by a
amcos(vtf), what is the value
of f? Make it positive, and if you cannot specify the value (such as
p/2 rad), then give a range of values (such as between 0 and p/2).
7Figure 15-24 shows
the x(t) curves for three
experiments involving a
particular spring–box
system oscillating in
SHM. Rank the curves
according to (a) the sys-
tem’s angular frequency,
(b) the spring’s poten-
tial energy at time t0,
(c) the box’s kinetic en-
ergy at t0, (d) the
box’s speed at t0, and (e) the box’s maximum kinetic energy, great-
est first.
8Figure 15-25 shows plots of the kinetic energy Kversus
position xfor three harmonic oscillators that have the same mass.
Rank the plots according to (a) the
corresponding spring constant and
(b) the corresponding period of the
oscillator,greatest first.
9Figure 15-26 shows three physical
pendulums consisting of identical uni-
form spheres of the same mass that
are rigidly connected by identical rods
of negligible mass. Each pendulum is
vertical and can pivot about suspen-
sion point O. Rank the pendulums ac-
cording to their period of oscillation,
greatest first.
10 You are to build the oscillation
transfer device shown in Fig. 15-27. It
consists of two spring–block systems
hanging from a flexible rod. When
the spring of system 1 is stretched
and then released, the resulting SHM
of system 1 at frequency f1oscillates
the rod. The rod then exerts a driving force on system 2, at the same
frequency f1. You can choose from four springs with spring constants
kof 1600, 1500,1400, and 1200 N/m, and four blocks with masses mof
800, 500, 400, and 200 kg. Mentally determine which spring should go
with which block in each of the two systems to maximize the ampli-
tude of oscillations in system 2.
Figure 15-21 Question 3.
2
13
4
5
6
7
8t
a
Figure 15-22 Question 5.
t
EDCBA
(a)
Figure 15-23 Question 6.
Figure 15-24 Question 7.
12
3
x
t
(b)
x
t = 0
xm
–xm0
t
EDCBA
(a)
(b)
x
t = 0
xm
–xm0
System 1 System 2
Rod
Figure 15-27 Question 10.
A B
x
t
A B
x
t
(a)
AB
x
t
(c)(b)
Figure 15-29 Question 12.
–xmxm
K
A
B
C
x
Figure 15-25 Question 8.
O O
O
(a) (b) (c)
Figure 15-26 Question 9.
Figure 15-28 Question 11.
d1
d2
436 CHAPTER 15 OSCILLATIONS
are attached to a block of mass 0.245 kg.What is the frequency of
oscillation on the frictionless floor?
•12 What is the phase constant for
the harmonic oscillator with the veloc-
ity function v(t) given in Fig. 15-32 if
the position function x(t) has the form
xxmcos(vtf)? The vertical axis
scale is set by vs4.0 cm/s.
•13 An oscillator consists of a
block of mass 0.500 kg connected to
a spring.When set into oscillation with amplitude 35.0 cm, the os-
cillator repeats its motion every 0.500 s. Find the (a) period, (b)
frequency, (c) angular frequency, (d) spring constant, (e) maxi-
mum speed, and (f) magnitude of the maximum force on the
block from the spring.
••14 A simple harmonic oscillator consists of a block of mass
2.00 kg attached to a spring of spring constant 100 N/m. When
t1.00 s, the position and velocity of the block are x0.129
m and v3.415 m/s. (a) What is the amplitude of the oscilla-
tions? What were the (b) position and (c) velocity of the block
at t0s?
••15 Two particles oscillate in simple harmonic motion along
a common straight-line segment of length A. Each particle has a pe-
riod of 1.5 s, but they differ in phase by p/6 rad. (a) How far apart
are they (in terms of A) 0.50 s after the lagging particle leaves one
end of the path? (b) Are they then moving in the same direction,
toward each other,or away from each other?
••16 Two particles execute simple harmonic motion of the same
amplitude and frequency along close parallel lines. They pass each
other moving in opposite directions each time their displacement
is half their amplitude.What is their phase difference?
••17 An oscillator consists of a block attached to a spring (k
400 N/m). At some time t, the position (measured from the sys-
tem’s equilibrium location), velocity, and acceleration of the block
are x0.100 m, v13.6 m/s, and a123 m/s2. Calculate (a) the
frequency of oscillation, (b) the mass of the block, and (c) the am-
plitude of the motion.
••18 At a certain harbor, the tides cause the ocean surface to
rise and fall a distance d(from highest level to lowest level) in
simple harmonic motion, with a period of 12.5 h. How long does
it take for the water to fall a distance 0.250dfrom its highest
level?
••19 A block rides on a piston (a squat cylindrical piece) that is
moving vertically with simple harmonic motion. (a) If the SHM
has period 1.0 s, at what amplitude of motion will the block and
piston separate? (b) If the piston has an amplitude of 5.0 cm,
what is the maximum frequency for which the block and piston
will be in contact continuously?
••20 Figure 15-33ais a partial graph of the position function
x(t) for a simple harmonic oscillator with an angular frequency of
ILW
SSM
SSM
Module 15-1 Simple Harmonic Motion
•1 An object undergoing simple harmonic motion takes 0.25 s to
travel from one point of zero velocity to the next such point. The
distance between those points is 36 cm. Calculate the (a) period,
(b) frequency, and (c) amplitude of the motion.
•2 A 0.12 kg body undergoes simple harmonic motion of ampli-
tude 8.5 cm and period 0.20 s. (a) What is the magnitude of the
maximum force acting on it? (b) If the oscillations are produced by
a spring,what is the spring constant?
•3 What is the maximum acceleration of a platform that
oscillates at amplitude 2.20 cm and frequency 6.60 Hz?
•4 An automobile can be considered to be mounted on four identical
springs as far as vertical oscillations are concerned.The springs of a cer-
tain car are adjusted so that the oscillations have a frequency of 3.00
Hz. (a) What is the spring constant of each spring if the mass of the car
is 1450 kg and the mass is evenly distributed over the springs? (b) What
will be the oscillation frequency if five passengers, averaging 73.0 kg
each,ride in the car with an even distribution of mass?
•5 In an electric shaver,the blade moves back and forth over
a distance of 2.0 mm in simple harmonic motion, with frequency
120 Hz. Find (a) the amplitude, (b) the maximum blade speed, and
(c) the magnitude of the maximum blade acceleration.
•6 A particle with a mass of 1.00 kg is oscillating with
simple harmonic motion with a period of 1.00 s and a maxi-
mum speed of 1.00 103m/s. Calculate (a) the angular frequency
and (b) the maximum displacement of the particle.
•7 A loudspeaker produces a musical sound by means of the
oscillation of a diaphragm whose amplitude is limited to 1.00 . (a)
At what frequency is the magnitude aof the diaphragm’s acceleration
equal to g? (b) For greater frequencies, is agreater than or less than g?
•8 What is the phase constant for
the harmonic oscillator with the po-
sition function x(t) given in Fig. 15-
30 if the position function has the
form xx
mcos( )? The ver-
tical axis scale is set by xs6.0 cm.
•9 The position function x
(6.0 m) cos[(3prad/s)tp/3 rad]
gives the simple harmonic motion
of a body. At t2.0 s, what are the
(a) displacement, (b) velocity, (c)
acceleration, and (d) phase of the motion? Also, what are the (e)
frequency and (f) period of the motion?
•10 An oscillating block–spring system takes 0.75 s to begin re-
peating its motion. Find (a) the pe-
riod, (b) the frequency in hertz, and
(c) the angular frequency in radians
per second.
•11 In Fig. 15-31, two identical
springs of spring constant 7580 N/m
vtf
mm
SSM
105
1020
SSM
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
xs
t
x (cm)
–xs
Figure 15-30 Problem 8.
m
Figure 15-31
Problems 11 and 21.
vs
t
v(cm/s)
–vs
Figure 15-32 Problem 12.
surface. The coefficient of static
friction between the two blocks is
0.40. What amplitude of simple har-
monic motion of the spring–blocks
system puts the smaller block on
the verge of slipping over the larger block?
Module 15-2 Energy in Simple Harmonic Motion
•27 When the displacement in SHM is one-half the ampli-
tude xm, what fraction of the total energy is (a) kinetic energy and
(b) potential energy? (c) At what displacement, in terms of the am-
plitude, is the energy of the system half kinetic energy and half po-
tential energy?
•28 Figure 15-38 gives the one-
dimensional potential energy well
for a 2.0 kg particle (the function
U(x) has the form bx2and the ver-
tical axis scale is set by Us2.0 J).
(a) If the particle passes through
the equilibrium position with a ve-
locity of 85 cm/s, will it be turned
back before it reaches x15 cm?
(b) If yes, at what position, and if
no, what is the speed of the parti-
cle at x15 cm?
•29 Find the mechanical energy of a block–spring system
with a spring constant of 1.3 N/cm and an amplitude of 2.4 cm.
•30 An oscillating block–spring system has a mechanical energy of
1.00 J, an amplitude of 10.0 cm, and a maximum speed of 1.20 m/s.
Find (a) the spring constant, (b) the mass of the block, and (c) the
frequency of oscillation.
•31 A 5.00 kg object on a horizontal frictionless surface is at-
tached to a spring with k1000 N/m. The object is displaced from
equilibrium 50.0 cm horizontally and given an initial velocity of 10.0
m/s back toward the equilibrium position.What are (a) the motion’s
frequency, (b) the initial potential energy of the block–spring sys-
tem, (c) the initial kinetic energy, and (d) the motion’s amplitude?
•32 Figure 15-39 shows the ki-
netic energy Kof a simple
harmonic oscillator versus its po-
sition x. The vertical axis scale is
set by Ks4.0 J. What is the
spring constant?
••33 A block of mass M5.4
kg, at rest on a horizontal
frictionless table, is attached to a
rigid support by a spring of con-
stant k6000 N/m. A bullet of
mass m9.5 g and velocity of
magnitude 630 m/s strikes and is
embedded in the block (Fig. 15-
40).Assuming the compression of
the spring is negligible until the
bullet is embedded, determine (a)
the speed of the block immedi-
ately after the collision and (b)
the amplitude of the resulting simple harmonic motion.
v
:
ILW
SSM
SSM
437
PROBLEMS
1.20 rad/s; Fig. 15-33bis a par-
tial graph of the corresponding
velocity function v(t). The ver-
tical axis scales are set by xs
5.0 cm and vs5.0 cm/s. What
is the phase constant of the
SHM if the position function
x(t) is in the general form x
xmcos(vtf)?
••21 In Fig. 15-31, two
springs are attached to a block
that can oscillate over a fric-
tionless floor. If the left spring
is removed, the block oscil-
lates at a frequency of 30 Hz.
If, instead, the spring on the
right is removed, the block os-
cillates at a frequency of 45
Hz. At what frequency does
the block oscillate with both
springs attached?
••22 Figure 15-34 shows
block 1 of mass 0.200 kg slid-
ing to the right over a friction-
less elevated surface at a
speed of 8.00 m/s. The block
undergoes an elastic collision
with stationary block 2, which
is attached to a spring of spring
constant 1208.5 N/m. (Assume
that the spring does not affect the collision.) After the collision,
block 2 oscillates in SHM with a period of 0.140 s, and block 1
slides off the opposite end of the elevated surface, landing a dis-
tance dfrom the base of that surface after falling height h4.90
m.What is the value of d?
••23 A block is on a horizontal surface (a shake
table) that is moving back and forth horizontally with simple har-
monic motion of frequency 2.0 Hz.The coefficient of static friction
between block and surface is 0.50. How great can the amplitude of
the SHM be if the block is not to slip along the surface?
•••24 In Fig. 15-35, two springs are
joined and connected to a block of
mass 0.245 kg that is set oscillating
over a frictionless floor.The springs
each have spring constant k
6430 N/m. What is the frequency of
the oscillations?
•••25 In Fig. 15-36, a block
weighing 14.0 N, which can slide
without friction on an incline at an-
gle 40.0, is connected to the
top of the incline by a massless
spring of unstretched length 0.450
m and spring constant 120 N/m. (a)
How far from the top of the incline
is the block’s equilibrium point? (b)
If the block is pulled slightly down the incline and released, what is
the period of the resulting oscillations?
•••26 In Fig. 15-37, two blocks (m1.8 kg and M10 kg) and
u
WWWSSM
ILW
a spring (k200 N/m) are ar-
ranged on a horizontal, frictionless
Figure 15-33 Problem 20.
xs
–xs
x (cm)
t
(a)
vs
–vs
v (cm/s)
t
(b)
Figure 15-34 Problem 22.
12k
h
d
mkk
Figure 15-35 Problem 24.
θ
k
Figure 15-36 Problem 25.
k
m
M
Figure 15-37 Problem 26.
Figure 15-38 Problem 28.
–20 –10 0 10 20
x (cm)
U (J)
Us
–12 –8 –4 0 4 8 12
x (cm)
K (J)
Ks
Figure 15-39 Problem 32.
k
mM
v
Figure 15-40 Problem 33.
438 CHAPTER 15 OSCILLATIONS
••34 In Fig. 15-41, block 2 of
mass 2.0 kg oscillates on the end of a
spring in SHM with a period of 20
ms. The block’s position is given by
x(1.0 cm) cos(vtp/2). Block 1
of mass 4.0 kg slides toward block 2
with a velocity of magnitude 6.0 m/s, directed along the spring’s
length. The two blocks undergo a completely inelastic collision at
time t5.0 ms. (The duration of the collision is much less than
the period of motion.) What is the amplitude of the SHM after the
collision?
••35 A 10 g particle undergoes SHM with an amplitude of 2.0 mm,
a maximum acceleration of magnitude 8.0 103m/s2, and an
unknown phase constant f.What are (a) the period of the motion,
(b) the maximum speed of the particle, and (c) the total mechani-
cal energy of the oscillator? What is the magnitude of the force on
the particle when the particle is at (d) its maximum displacement
and (e) half its maximum displacement?
••36 If the phase angle for a block–spring system in SHM is p/6
rad and the block’s position is given by xxmcos(vtf), what is
the ratio of the kinetic energy to the potential energy at time t0?
•••37 A massless spring hangs from the ceiling with a small ob-
ject attached to its lower end.The object is initially held at rest in a
position yisuch that the spring is at its rest length. The object is
then released from yiand oscillates up and down, with its lowest
position being 10 cm below yi. (a) What is the frequency of the os-
cillation? (b) What is the speed of the object when it is 8.0 cm be-
low the initial position? (c) An object of mass 300 g is attached to
the first object, after which the system oscillates with half the origi-
nal frequency. What is the mass of the first object? (d) How far be-
low yiis the new equilibrium (rest) position with both objects at-
tached to the spring?
Module 15-3 An Angular Simple Harmonic Oscillator
•38 A 95 kg solid sphere with a 15 cm radius is suspended by a
vertical wire.A torque of 0.20 N m is required to rotate the sphere
through an angle of 0.85 rad and then maintain that orientation.
What is the period of the oscillations that result when the sphere is
then released?
••39 The balance wheel of an old-fashioned watch
oscillates with angular amplitude prad and period 0.500 s. Find
(a) the maximum angular speed of the wheel, (b) the angular speed
at displacement p/2 rad, and (c) the magnitude of the angular
acceleration at displacement p/4 rad.
Module 15-4 Pendulums, Circular Motion
•40 A physical pendulum consists of a meter stick that is piv-
oted at a small hole drilled through
the stick a distance dfrom the 50
cm mark. The period of oscillation
is 2.5 s. Find d.
•41 In Fig. 15-42, the pendu-
lum consists of a uniform disk with
radius r10.0 cm and mass 500 g
attached to a uniform rod with
length L500 mm and mass 270
g. (a) Calculate the rotational iner-
tia of the pendulum about the
pivot point. (b) What is the dis-
tance between the pivot point and
SSM
ILW
WWWSSM
the center of mass of the pendulum? (c) Calculate the period of
oscillation.
•42 Suppose that a simple pendulum consists of a small 60.0 g
bob at the end of a cord of negligible mass. If the angle ubetween
the cord and the vertical is given by
u(0.0800 rad) cos[(4.43 rad/s)tf],
what are (a) the pendulum’s length and (b) its maximum kinetic
energy?
•43 (a) If the physical pendulum of Fig. 15-13 and the associated
sample problem is inverted and suspended at point P, what is its
period of oscillation? (b) Is the period now greater than, less than,
or equal to its previous value?
•44 A physical pendulum consists of two me-
ter-long sticks joined together as shown in Fig.
15-43. What is the pendulum’s period of oscilla-
tion about a pin inserted through point Aat the
center of the horizontal stick?
•45 A performer seated on a trapeze is
swinging back and forth with a period of 8.85 s.
If she stands up, thus raising the center of mass of
the trapeze performer system by 35.0 cm, what will be the new pe-
riod of the system? Treat trapeze performer as a simple pendulum.
•46 A physical pendulum has a center of oscillation at distance
2L/3 from its point of suspension. Show that the distance be-
tween the point of suspension and the center of oscillation for a
physical pendulum of any form is I/mh, where Iand hhave the
meanings in Eq. 15-29 and mis the mass of the pendulum.
•47 In Fig. 15-44, a physical pendulum
consists of a uniform solid disk (of radius
R2.35 cm) supported in a vertical plane
by a pivot located a distance d1.75 cm
from the center of the disk.The disk is dis-
placed by a small angle and released.
What is the period of the resulting simple
harmonic motion?
••48 A rectangular block, with face
1 2
k
Figure 15-41 Problem 34.
L
r
Figure 15-42 Problem 41.
A
Figure 15-43
Problem 44.
Pivot
d
R
lengths a35 cm and b45 cm, is to be
suspended on a thin horizontal rod running through a narrow hole in
the block. The block is then to be set swinging about the rod like a
pendulum, through small angles so that it is in SHM. Figure 15-45
shows one possible position of the hole, at distance rfrom the block’s
center, along a line connecting the center with a corner. (a) Plot the
period versus distance ralong that
line such that the minimum in the
curve is apparent. (b) For what value
of rdoes that minimum occur? There
is a line of points around the block’s
center for which the period of swing-
ing has the same minimum value. (c)
What shape does that line make?
••49 The angle of the pendulum
of Fig. 15-11bis given by u
umcos[(4.44 rad/s)tf]. If at t0,
u0.040 rad and du/dt 0.200
rad/s, what are (a) the phase con-
Figure 15-44
Problem 47.
Figure 15-45 Problem 48.
a
b
r
stant fand (b) the maximum angle um? (Hint: Don’t confuse the
rate du/dt at which uchanges with the vof the SHM.)
lengths d6.00 cm and is mounted on an axle
through its center. A spring (k1200 N/m) con-
nects the cube’s upper corner to a rigid wall.
Initially the spring is at its rest length. If the cube
is rotated 3and released, what is the period of
the resulting SHM?
••53 In the overhead view of Fig. 15-
48, a long uniform rod of mass 0.600 kg is free to
rotate in a horizontal plane about a
vertical axis through its center. A
spring with force constant k1850
N/m is connected horizontally be-
tween one end of the rod and a
fixed wall. When the rod is in equi-
librium, it is parallel to the wall.
What is the period of the small os-
cillations that result when the rod is rotated slightly and released?
••54 In Fig. 15-49a, a metal plate is mounted on an axle through
its center of mass.A spring with k2000 N/m connects a wall with a
point on the rim a distance r2.5 cm from the center of mass.
Initially the spring is at its rest length. If the plate is rotated by 7and
released, it rotates about the axle in SHM, with its angular position
given by Fig.15-49b.The horizontal axis scale is set by ts20 ms.What
is the rotational inertia of the plate about its center of mass?
ILWSSM
gives the least period? (b) What is
that least period?
••52 The 3.00 kg cube in Fig. 15-47 has edge
length L1.85 m oscillates as a
physical pendulum. (a) What value
of distance xbetween the stick’s
center of mass and its pivot point O
••51 In Fig. 15-46, a stick of
439
PROBLEMS
••50 A thin uniform rod (mass 0.50 kg) swings about an
axis that passes through one end of the rod and is perpendicu-
lar to the plane of the swing. The
rod swings with a period of 1.5 s
and an angular amplitude of 10.
(a) What is the length of the rod?
(b) What is the maximum kinetic
energy of the rod as it swings?
sen to minimize the period and then Lis increased, does the pe-
riod increase, decrease, or remain the same? (c) If, instead, mis in-
creased without Lincreasing, does
the period increase, decrease, or re-
main the same?
L/2
O
L/2
x
Figure 15-46 Problem 51.
•••56 In Fig. 15-50, a 2.50 kg disk
of diameter D42.0 cm is sup-
ported by a rod of length L76.0
cm and negligible mass that is piv-
oted at its end. (a) With the massless
torsion spring unconnected, what is
the period of oscillation? (b) With
the torsion spring connected, the rod
is vertical at equilibrium.What is the
torsion constant of the spring if the
period of oscillation has been de-
creased by 0.500 s?
Module 15-5 Damped Simple Harmonic Motion
•57 The amplitude of a lightly damped oscillator decreases by
3.0% during each cycle.What percentage of the mechanical energy
of the oscillator is lost in each cycle?
•58 For the damped oscillator system shown in Fig. 15-16, with
m 250 g, k 85 N/m, and b70 g/s, what is the ratio of the oscil-
lation amplitude at the end of 20 cycles to the initial oscillation
amplitude?
•59 For the damped oscillator system shown in Fig.
15-16, the block has a mass of 1.50 kg and the spring constant is
8.00 N/m.The damping force is given by b(dx/dt), where b230
g/s. The block is pulled down 12.0 cm and released. (a) Calculate
the time required for the amplitude of the resulting oscillations to
fall to one-third of its initial value. (b) How many oscillations are
made by the block in this time?
••60 The suspension system of a 2000 kg automobile “sags” 10 cm
when the chassis is placed on it. Also, the oscillation amplitude
decreases by 50% each cycle. Estimate the values of (a) the
spring constant kand (b) the damping constant bfor the spring
and shock absorber system of one wheel, assuming each wheel
supports 500 kg.
Module 15-6 Forced Oscillations and Resonance
•61 For Eq. 15-45, suppose the amplitude xmis given by
where Fmis the (constant) amplitude of the external oscillating
force exerted on the spring by the rigid support in Fig. 15-16. At
resonance, what are the (a) amplitude and (b) velocity amplitude
of the oscillating object?
•62 Hanging from a horizontal beam are nine simple pendulums
of the following lengths: (a) 0.10, (b) 0.30, (c) 0.40, (d) 0.80, (e) 1.2,
(f) 2.8, (g) 3.5, (h) 5.0, and (i) 6.2 m. Suppose the beam undergoes
horizontal oscillations with angular frequencies in the range from
2.00 rad/s to 4.00 rad/s. Which of the pendulums will be (strongly)
set in motion?
••63 A 1000 kg car carrying four 82 kg people travels over a
“washboard” dirt road with corrugations 4.0 m apart. The car
bounces with maximum amplitude when its speed is 16 km/h.
When the car stops, and the people get out, by how much does the
car body rise on its suspension?
xmFm
[m2(v2
dv2)2b2v2
d]1/2 ,
WWWSSM
d
d
Figure 15-47
Problem 52.
r
(a) (b)
ts
8
4
00
–4
–8
t(ms)
θ
(deg)
Figure 15-49 Problem 54.
k
Rotation axis
Wall
Figure 15-48 Problem 53.
•••55 A pendulum is formed by pivoting a long thin rod
about a point on the rod. In a series of experiments, the period is
measured as a function of the distance xbetween the pivot point
and the rod’s center. (a) If the rod’s length is L2.20 m and its
mass is m22.1 g, what is the minimum period? (b) If xis cho-
Figure 15-50 Problem 56.
D
L
440 CHAPTER 15 OSCILLATIONS
Figure 15-53 Problems 77 and 78.
x (cm)
8
4
–4
–8
0ts
t (ms)
Additional Problems
64 Although California is known for earthquakes, it has
large regions dotted with precariously balanced rocks that would
be easily toppled by even a mild earthquake. Apparently no major
earthquakes have occurred in those regions. If an earthquake were
to put such a rock into sinusoidal oscillation (parallel to the
ground) with a frequency of 2.2 Hz, an oscillation amplitude of 1.0
cm would cause the rock to topple. What would be the magnitude
of the maximum acceleration of the oscillation, in terms of g?
65 A loudspeaker diaphragm is oscillating in simple harmonic
motion with a frequency of 440 Hz and a maximum displacement
of 0.75 mm. What are the (a) angular frequency, (b) maximum
speed, and (c) magnitude of the maximum acceleration?
66 A uniform spring with k8600 N/m is cut into pieces 1 and 2
of unstretched lengths L17.0 cm and L210 cm. What are
(a) k1and (b) k2? A block attached to the original spring as in
Fig. 15-7 oscillates at 200 Hz. What is the oscillation frequency of
the block attached to (c) piece 1 and (d) piece 2?
67 In Fig. 15-51, three 10 000 kg
ore cars are held at rest on a mine
railway using a cable that is parallel
to the rails, which are inclined at an-
gle u30. The cable stretches 15
cm just before the coupling between
the two lower cars breaks, detaching
the lowest car.Assuming that the ca-
ble obeys Hooke’s law, find the (a)
frequency and (b) amplitude of the
resulting oscillations of the remain-
ing two cars.
68 A 2.00 kg block hangs from a
spring.A 300 g body hung below the block stretches the spring 2.00 cm
farther.(a) What is the spring constant? (b) If the 300 g body is removed
and the block is set into oscillation,find the period of the motion.
69 In the engine of a locomotive, a cylindrical piece known
as a piston oscillates in SHM in a cylinder head (cylindrical cham-
ber) with an angular frequency of 180 rev/min. Its stroke (twice the
amplitude) is 0.76 m.What is its maximum speed?
70 A wheel is free to rotate about its fixed axle. A spring is at-
tached to one of its spokes a distance rfrom the axle, as shown in Fig.
15-52. (a) Assuming that the wheel is a hoop of mass mand radius R,
what is the angular frequency vof
small oscillations of this system in
terms of m,R,r, and the spring con-
stant k? What is vif (b) rRand (c)
r0?
71 A 50.0 g stone is attached to
the bottom of a vertical spring and
set vibrating. If the maximum speed
of the stone is 15.0 cm/s and the pe-
riod is 0.500 s, find the (a) spring
constant of the spring, (b) amplitude of the motion, and (c) fre-
quency of oscillation.
72 A uniform circular disk whose radius Ris 12.6 cm is suspended
as a physical pendulum from a point on its rim. (a) What is its pe-
riod? (b) At what radial distance rRis there a pivot point that
gives the same period?
73 A vertical spring stretches 9.6 cm when a 1.3 kg block
SSM
SSM
is hung from its end. (a) Calculate the spring constant.This block
is then displaced an additional 5.0 cm downward and released
from rest. Find the (b) period, (c) frequency, (d) amplitude, and
(e) maximum speed of the resulting SHM.
74 A massless spring with spring constant 19 N/m hangs vertically.
A body of mass 0.20 kg is attached to its free end and then released.
Assume that the spring was unstretched before the body was re-
leased. Find (a) how far below the initial position the body descends,
and the (b) frequency and (c) amplitude of the resulting SHM.
75 A 4.00 kg block is suspended from a spring with k500 N/m.A
50.0 g bullet is fired into the block from directly below with a speed of
150 m/s and becomes embedded in the block. (a) Find the amplitude
of the resulting SHM. (b) What percentage of the original kinetic en-
ergy of the bullet is transferred to mechanical energy of the oscillator?
76 A 55.0 g block oscillates in SHM on the end of a spring with
k1500 N/m according to xxmcos(vtf). How long does
the block take to move from position 0.800xmto (a) position
0.600xmand (b) position 0.800xm?
77 Figure 15-53 gives the position of a 20 g block oscillating
in SHM on the end of a spring. The horizontal axis scale is set
by ts40.0 ms. What are (a) the maximum kinetic energy of the
block and (b) the number of times per second that maximum is
reached? (Hint: Measuring a slope will probably not be very ac-
curate. Find another approach.)
78 Figure 15-53 gives the position x(t) of a block oscillating in SHM
on the end of a spring (ts40.0 ms).What are (a) the speed and (b) the
magnitude of the radial acceleration of a particle in the corresponding
uniform circular motion?
79 Figure 15-54 shows the kinetic
energy Kof a simple pendulum versus
its angle ufrom the vertical.The verti-
cal axis scale is set by Ks10.0 mJ.
The pendulum bob has mass 0.200 kg.
What is the length of the pendulum?
80 A block is in SHM on the end
of a spring, with position given by
xxmcos(vtf). If fp/5 rad,
then at t0 what percentage of the
total mechanical energy is potential energy?
81 A simple harmonic oscillator consists of a 0.50 kg block at-
tached to a spring. The block slides back and forth along a straight
line on a frictionless surface with equilibrium point x0. At t0
the block is at x0 and moving in the positive xdirection.A graph
of the magnitude of the net force on the block as a function of itsF
:
Figure 15-51 Problem 67.
θ
Car that
breaks free
k
rR
Figure 15-52 Problem 70.
–
100 –50 0
(mrad)
K (mJ)
θ
50 100
Ks
Figure 15-54 Problem 79.
441
PROBLEMS
position is shown in Fig. 15-55.
The vertical scale is set by Fs
75.0 N. What are (a) the ampli-
tude and (b) the period of the
motion, (c) the magnitude of the
maximum acceleration, and (d)
the maximum kinetic energy?
82 A simple pendulum of
length 20 cm and mass 5.0 g is
suspended in a race car traveling with constant speed 70 m/s
around a circle of radius 50 m. If the pendulum undergoes small
oscillations in a radial direction about its equilibrium position,
what is the frequency of oscillation?
83 The scale of a spring balance that reads from 0 to 15.0 kg is
12.0 cm long. A package suspended from the balance is found to
oscillate vertically with a frequency of 2.00 Hz. (a) What is the
spring constant? (b) How much does the package weigh?
84 A 0.10 kg block oscillates back and forth along a straight line
on a frictionless horizontal surface. Its displacement from the ori-
gin is given by
x(10 cm) cos[(10 rad/s)tp/2 rad].
(a) What is the oscillation frequency? (b) What is the maxi-
mum speed acquired by the block? (c) At what value of xdoes
this occur? (d) What is the magnitude of the maximum accel-
eration of the block? (e) At what value of xdoes this occur?
(f) What force, applied to the block by the spring, results in the
given oscillation?
85 The end point of a spring oscillates with a period of 2.0 s when
a block with mass mis attached to it. When this mass is increased
by 2.0 kg, the period is found to be 3.0 s. Find m.
86 The tip of one prong of a tuning fork undergoes SHM of fre-
quency 1000 Hz and amplitude 0.40 mm. For this tip, what is the
magnitude of the (a) maximum acceleration, (b) maximum veloc-
ity, (c) acceleration at tip displacement 0.20 mm, and (d) velocity at
tip displacement 0.20 mm?
87 A flat uniform circular disk has a mass of 3.00 kg and a radius
of 70.0 cm. It is suspended in a horizontal plane by a vertical wire at-
tached to its center. If the disk is rotated 2.50 rad about the wire, a
torque of 0.0600 N m is required to maintain that orientation.
Calculate (a) the rotational inertia of the disk about the wire, (b) the
torsion constant, and (c) the angular frequency of this torsion pendu-
lum when it is set oscillating.
88 A block weighing 20 N oscillates at one end of a vertical
spring for which k100 N/m; the other end of the spring is at-
tached to a ceiling. At a certain instant the spring is stretched
0.30 m beyond its relaxed length (the length when no object is at-
tached) and the block has zero velocity. (a) What is the net force on
the block at this instant? What are the (b) amplitude and (c) period
of the resulting simple harmonic motion? (d) What is the maxi-
mum kinetic energy of the block as it oscillates?
89 A 3.0 kg particle is in simple harmonic motion in one
dimension and moves according to the equation
x(5.0 m) cos[(p/3 rad/s)tp/4 rad],
with tin seconds. (a) At what value of xis the potential energy of the
particle equal to half the total energy? (b) How long does the parti-
cle take to move to this position xfrom the equilibrium position?
90 A particle executes linear SHM with frequency 0.25 Hz about
the point x0. At t0, it has displacement x0.37 cm and zero
velocity. For the motion, determine the (a) period, (b) angular
frequency, (c) amplitude, (d) displacement x(t), (e) velocity v(t),
(f) maximum speed, (g) magnitude of the maximum acceleration,
(h) displacement at t3.0 s, and (i) speed at t3.0 s.
91 What is the frequency of a simple pendulum 2.0 m long
(a) in a room, (b) in an elevator accelerating upward at a rate of
2.0 m/s2,and (c) in free fall?
92 A grandfather clock has a pen-
dulum that consists of a thin brass disk
of radius r15.00 cm and mass 1.000
kg that is attached to a long thin rod of
negligible mass. The pendulum swings
freely about an axis perpendicular to
the rod and through the end of the rod
opposite the disk, as shown in Fig.
15-56. If the pendulum is to have a pe-
riod of 2.000 s for small oscillations at a
place where g9.800 m/s2, what must
be the rod length Lto the nearest tenth
of a millimeter?
93 A 4.00 kg block hangs from a
spring, extending it 16.0 cm from its
unstretched position. (a) What is the spring constant? (b) The
block is removed, and a 0.500 kg body is hung from the same
spring. If the spring is then stretched and released, what is its pe-
riod of oscillation?
94 What is the phase constant for
SMH with a(t) given in Fig. 15-57 if
the position function x(t) has the form
xxmcos(vtf) and as4.0 m/s2?
95 An engineer has an odd-shaped
10 kg object and needs to find its rota-
tional inertia about an axis through its
center of mass.The object is supported
on a wire stretched along the desired
axis. The wire has a torsion constant
k0.50 Nm. If this torsion pendulum oscillates through 20 cycles in
50 s, what is the rotational inertia of the object?
96 A spider can tell when its web has captured, say, a fly
because the fly’s thrashing causes the web threads to oscillate.A
spider can even determine the size of the fly by the frequency of
the oscillations.Assume that
a fly oscillates on the cap-
ture thread on which it is
caught like a block on a
spring. What is the ratio of
oscillation frequency for a
fly with mass mto a fly with
mass 2.5m?
97 A torsion pendulum
consists of a metal disk with
a wire running through its
center and soldered in place.
The wire is mounted verti-
cally on clamps and pulled
taut. Figure 15-58agives the
magnitude tof the torque
SSM
Figure 15-55 Problem 81.
F (N)
x (m)
–0.30
0.30
–Fs
Fs
Figure 15-56 Problem 92.
Rotation
axis
L
r
as
–as
t
a (m/s2)
Figure 15-57 Problem 94.
(10–3 N m)
τ
0.100
(rad)
(a)
(b)
0.20
θ
θ
(rad)
0.2
0
–0.2
0ts
t(s)
τ
s
Figure 15-58 Problem 97.
face is attached to a horizontal spring with k480 N/m. Let xbe
the displacement of the block from the position at which the spring
is unstretched. At t0 the block passes through x0 with a
speed of 5.2 m/s in the positive xdirection. What are the (a) fre-
quency and (b) amplitude of the block’s motion? (c) Write an ex-
pression for xas a function of time.
102 A simple harmonic oscillator consists of an 0.80 kg block at-
tached to a spring (k200 N/m). The block slides on a horizontal
frictionless surface about the equilibrium point x0 with a total
mechanical energy of 4.0 J. (a) What is the amplitude of the oscilla-
tion? (b) How many oscillations does the block complete in 10 s?
(c) What is the maximum kinetic energy attained by the block? (d)
What is the speed of the block at x0.15 m?
103 A block sliding on a horizontal frictionless surface is
attached to a horizontal spring with a spring constant of 600 N/m.
The block executes SHM about its equilibrium position with a pe-
riod of 0.40 s and an amplitude of 0.20 m. As the block slides
through its equilibrium position, a 0.50 kg putty wad is dropped
442 CHAPTER 15 OSCILLATIONS
needed to rotate the disk about its center (and thus twist the
wire) versus the rotation angle u. The vertical axis scale is set by
ts4.0 103Nm.The disk is rotated to u0.200 rad and then
released. Figure 15-58bshows the resulting oscillation in terms of
angular position uversus time t. The horizontal axis scale is set by
ts0.40 s. (a) What is the rotational inertia of the disk about
its center? (b) What is the maximum angular speed du/dt of the
disk? (Caution: Do not confuse the (constant) angular frequency
of the SHM with the (varying) angular speed of the rotating disk,
even though they usually have the same symbol v.Hint: The po-
tential energy Uof a torsion pendulum is equal to ku2, analogous
to Ukx2for a spring.)
98 When a 20 N can is hung from the bottom of a vertical spring, it
causes the spring to stretch 20 cm. (a) What is the spring constant? (b)
This spring is now placed horizontally on a frictionless table. One end
of it is held fixed,and the other end is attached to a 5.0 N can.The can is
then moved (stretching the spring) and released from rest.What is the
period of the resulting oscillation?
99 For a simple pendulum, find the angular amplitude umat
which the restoring torque required for simple harmonic motion
deviates from the actual restoring torque by 1.0%. (See
“Trigonometric Expansions” in Appendix E.)
100 In Fig. 15-59, a solid cylinder
attached to a horizontal spring (k
3.00 N/m) rolls without slipping
along a horizontal surface. If the sys-
tem is released from rest when the
spring is stretched by 0.250 m, find
(a) the translational kinetic energy
and (b) the rotational kinetic energy of the cylinder as it passes
through the equilibrium position. (c) Show that under these condi-
tions the cylinder’s center of mass executes simple harmonic mo-
tion with period
T2p
where Mis the cylinder mass. (Hint: Find the time derivative of the
total mechanical energy.)
101 A 1.2 kg block sliding on a horizontal frictionless sur-
SSM
A3M
2k,
1
2
1
2
vertically onto the block. If the putty wad sticks to the block, de-
termine (a) the new period of the motion and (b) the new ampli-
tude of the motion.
104 A damped harmonic oscillator consists of a block (m
2.00 kg), a spring (k10.0 N/m), and a damping force (Fbv).
Initially, it oscillates with an amplitude of 25.0 cm; because of
the damping, the amplitude falls to three-fourths of this initial
value at the completion of four oscillations. (a) What is the
value of b? (b) How much energy has been “lost” during these
four oscillations?
105 A block weighing 10.0 N is attached to the lower end of a
vertical spring (k200.0 N/m), the other end of which is attached
to a ceiling.The block oscillates vertically and has a kinetic energy
of 2.00 J as it passes through the point at which the spring is
unstretched. (a) What is the period of the oscillation? (b) Use the
law of conservation of energy to determine the maximum distance
the block moves both above and below the point at which the
spring is unstretched. (These are not necessarily the same.)
(c) What is the amplitude of the oscillation? (d) What is the maxi-
mum kinetic energy of the block as it oscillates?
106 A simple harmonic oscillator consists of a block attached
to a spring with k200 N/m. The block slides on a frictionless
surface, with equilibrium point x0 and amplitude 0.20 m.
A graph of the block’s velocity vas a function of time tis shown
in Fig. 15-60.The horizontal scale is set by ts0.20 s.What are (a)
the period of the SHM, (b) the block’s mass, (c) its displacement
at t0, (d) its acceleration at t0.10 s, and (e) its maximum ki-
netic energy?
Mk
Figure 15-59 Problem 100.
v (m/s)
2
0
π
–2
π
t (s)
ts
Figure 15-60 Problem 106.
h
k
k
Figure 15-61 Problem 108.
107 The vibration frequencies of atoms in solids at normal temper-
atures are of the order of 1013 Hz. Imagine the atoms to be connected
to one another by springs. Suppose that a single silver atom in a solid
vibrates with this frequency and that all the other atoms are at rest.
Compute the effective spring constant. One mole of silver (6.02
1023 atoms) has a mass of 108 g.
108 Figure 15-61 shows that if we hang a block on the end of a
spring with spring constant k, the spring is stretched by distance
h2.0 cm. If we pull down on the block a short distance and
then release it, it oscillates vertically with a certain frequency.
What length must a simple pendulum have to swing with that
frequency?
443
PROBLEMS
109 The physical pendulum in Fig.
15-62 has two possible pivot points A
and B. Point Ahas a fixed position
but Bis adjustable along the length
of the pendulum as indicated by the
scaling.When suspended from A, the
pendulum has a period of T1.80 s.
The pendulum is then suspended
from B, which is moved until the
pendulum again has that period.
What is the distance Lbetween A
and B?
110 A common device for enter-
taining a toddler is a jump seat that
hangs from the horizontal portion of a
doorframe via elastic cords (Fig. 15-63).
Assume that only one cord is on each
side in spite of the more realistic
arrangement shown. When a child is
placed in the seat, they both descend by a distance dsas the cords
stretch (treat them as springs).Then the seat is pulled down an ex-
tra distance dmand released, so that the child oscillates vertically,
like a block on the end of a spring.Suppose you are the safety engi-
neer for the manufacturer of the seat. You do not want the magni-
tude of the child’s acceleration to exceed 0.20gfor fear of hurting
the child’s neck. If dm10 cm, what value of dscorresponds to that
acceleration magnitude?
112 In Fig. 15-64, a
2500 kg demolition
ball swings from the
end of a crane. The
length of the swinging
segment of cable is
17 m. (a) Find the pe-
riod of the swinging,
assuming that the sys-
tem can be treated as
a simple pendulum.
(b) Does the period
depend on the ball’s
mass?
113 The cen-
ter of oscillation of a
physical pendulum
has this interesting property: If an impulse (assumed horizontal
and in the plane of oscillation) acts at the center of oscillation, no
oscillations are felt at the point of support. Baseball players (and
players of many other sports) know that unless the ball hits the bat
at this point (called the “sweet spot” by athletes), the oscillations
due to the impact will sting their hands. To prove this property, let
the stick in Fig. 15-13asimulate a baseball bat. Suppose that a hori-
zontal force (due to impact with the ball) acts toward the right
at P, the center of oscillation. The batter is assumed to hold the
bat at O, the pivot point of the stick. (a) What acceleration does the
point Oundergo as a result of ? (b) What angular acceleration is
produced by about the center of mass of the stick? (c) As a re-
sult of the angular acceleration in (b), what linear acceleration
does point Oundergo? (d) Considering the magnitudes and direc-
tions of the accelerations in (a) and (c), convince yourself that Pis
indeed the “sweet spot.”
114 A (hypothetical) large slingshot is stretched 2.30 m to
launch a 170 g projectile with speed sufficient to escape from
Earth (11.2 km/s). Assume the elastic bands of the slingshot obey
Hooke’s law. (a) What is the spring constant of the device if all the
elastic potential energy is converted to kinetic energy? (b)
Assume that an average person can exert a force of 490 N. How
many people are required to stretch the elastic bands?
115 What is the length of a simple pendulum whose full swing
from left to right and then back again takes 3.2 s?
116 A 2.0 kg block is attached to the end of a spring with a spring
constant of 350 N/m and forced to oscillate by an applied force F
(15 N) sin(vdt), where vd35 rad/s. The damping constant is b
15 kg/s.At t0, the block is at rest with the spring at its rest length.
(a) Use numerical integration to plot the displacement of the block
for the first 1.0 s.Use the motion near the end of the 1.0 s interval to
estimate the amplitude, period, and angular frequency. Repeat the
calculation for (b) and (c) vd20 rad/s.vd1k/m
F
:F
:
F
:
Figure 15-63 Problem 110.
Figure 15-64 Problem 112.
L
com
B
A
Figure 15-62 Problem 109.
111 A 2.0 kg block executes SHM while attached to a horizontal
spring of spring constant 200 N/m. The maximum speed of the
block as it slides on a horizontal frictionless surface is 3.0 m/s.What
are (a) the amplitude of the block’s motion, (b) the magnitude of
its maximum acceleration, and (c) the magnitude of its minimum
acceleration? (d) How long does the block take to complete 7.0 cy-
cles of its motion?
CHAPTER 16
Waves—I
16-1 TRANSVERSE WAVES
After reading this module, you should be able to . . .
16.01 Identify the three main types of waves.
16.02 Distinguish between transverse waves and longitudi-
nal waves.
16.03 Given a displacement function for a traverse wave,
determine amplitude , angular wave number k, angular
frequency v, phase constant f, and direction of travel,
and calculate the phase kx vtfand the displace-
ment at any given time and position.
16.04 Given a displacement function for a traverse
wave, calculate the time between two given displace-
ments.
16.05 Sketch a graph of a transverse wave as a function
of position, identifying amplitude , wavelength l, where
the slope is greatest, where it is zero, and where the
string elements have positive velocity, negative velocity,
and zero velocity.
16.06 Given a graph of displacement versus time for
a transverse wave, determine amplitude and
period T.
ym
ym
ym
16.07 Describe the effect on a transverse wave of changing
phase constant f.
16.08 Apply the relation between the wave speed v, the
distance traveled by the wave, and the time required for
that travel.
16.09 Apply the relationships between wave speed v,
angular frequency v, angular wave number k, wavelength
l, period T, and frequency f.
16.10 Describe the motion of a string element as a trans-
verse wave moves through its location, and identify
when its transverse speed is zero and when it is maxi-
mum.
16.11 Calculate the transverse velocity u(t) of a string
element as a transverse wave moves through its location.
16.12 Calculate the transverse acceleration a(t) of a
string element as a transverse wave moves through its
location.
16.13 Given a graph of displacement, transverse velocity,
or transverse acceleration, determine the phase con-
stant f.
Key Ideas
Learning Objectives
444
●Mechanical waves can exist only in material media and are
governed by Newton’s laws. Transverse mechanical waves,
like those on a stretched string, are waves in which the
particles of the medium oscillate perpendicular to the wave’s
direction of travel. Waves in which the particles of the
medium oscillate parallel to the wave’s direction of travel are
longitudinal waves.
●A sinusoidal wave moving in the positive direction of an
xaxis has the mathematical form
y(x,t)ymsin(kx vt),
where ymis the amplitude (magnitude of the maximum dis-
placement) of the wave, kis the angular wave number, vis
the angular frequency, and kx vtis the phase. The wave-
length lis related to kby
k2p
l.
●The period Tand frequency fof the wave are related to vby
●The wave speed v(the speed of the wave along the string) is
related to these other parameters by
●Any function of the form
y(x,t)h(kx vt)
can represent a traveling wave with a wave speed as given
above and a wave shape given by the mathematical form of h.
The plus sign denotes a wave traveling in the negative
direction of the xaxis, and the minus sign a wave traveling in
the positive direction.
vv
kl
Tlf.
v
2pf1
T.
445
16-1 TRANSVERSE WAVES
What Is Physics?
One of the primary subjects of physics is waves. To see how important waves are
in the modern world, just consider the music industry. Every piece of music you
hear, from some retro-punk band playing in a campus dive to the most eloquent
concerto playing on the web, depends on performers producing waves and your
detecting those waves. In between production and detection, the information
carried by the waves might need to be transmitted (as in a live performance on
the web) or recorded and then reproduced (as with CDs, DVDs, or the other
devices currently being developed in engineering labs worldwide). The
financial importance of controlling music waves is staggering, and the rewards to
engineers who develop new control techniques can be rich.
This chapter focuses on waves traveling along a stretched string, such as on
a guitar. The next chapter focuses on sound waves, such as those produced by
a guitar string being played. Before we do all this, though, our first job is to
classify the countless waves of the everyday world into basic types.
Types of Waves
Waves are of three main types:
1. Mechanical waves. These waves are most familiar because we encounter them
almost constantly; common examples include water waves, sound waves, and
seismic waves. All these waves have two central features: They are governed
by Newton’s laws, and they can exist only within a material medium, such as
water,air,and rock.
2. Electromagnetic waves. These waves are less familiar, but you use them
constantly; common examples include visible and ultraviolet light, radio and
television waves, microwaves, x rays, and radar waves. These waves require no
material medium to exist. Light waves from stars, for example, travel through
the vacuum of space to reach us. All electromagnetic waves travel through a
vacuum at the same speed c299 792 458 m/s.
3. Matter waves. Although these waves are commonly used in modern technol-
ogy, they are probably very unfamiliar to you. These waves are associated
with electrons, protons, and other fundamental particles, and even atoms and
molecules. Because we commonly think of these particles as constituting
matter,such waves are called matter waves.
Much of what we discuss in this chapter applies to waves of all kinds.
However,for specific examples we shall refer to mechanical waves.
Transverse and Longitudinal Waves
A wave sent along a stretched, taut string is the simplest mechanical wave. If you
give one end of a stretched string a single up-and-down jerk, a wave in the form
of a single pulse travels along the string. This pulse and its motion can occur
because the string is under tension.When you pull your end of the string upward,
it begins to pull upward on the adjacent section of the string via tension between
the two sections. As the adjacent section moves upward, it begins to pull the next
section upward, and so on. Meanwhile, you have pulled down on your end of the
string. As each section moves upward in turn, it begins to be pulled back
downward by neighboring sections that are already on the way down. The net
result is that a distortion in the string’s shape (a pulse, as in Fig. 16-1a) moves
along the string at some velocity .v
:
Figure 16-1 (a) A single pulse is sent along
a stretched string.A typical string element
(marked with a dot) moves up once and
then down as the pulse passes.The ele-
ment’s motion is perpendicular to the
wave’s direction of travel,so the pulse is a
transverse wave.(b) A sinusoidal wave is
sent along the string.A typical string
element moves up and down continuously
as the wave passes.This too is a transverse
wave.
y
x
y
x
(a)
(b)
Sinusoidal
wave
Pulse
v
v
446 CHAPTER 16 WAVES—I
If you move your hand up and down in continuous simple harmonic motion, a
continuous wave travels along the string at velocity . Because the motion of your
hand is a sinusoidal function of time, the wave has a sinusoidal shape at any given in-
stant, as in Fig. 16-1b; that is, the wave has the shape of a sine curve or a cosine curve.
We consider here only an “ideal” string, in which no friction-like forces
within the string cause the wave to die out as it travels along the string. In
addition, we assume that the string is so long that we need not consider a wave
rebounding from the far end.
One way to study the waves of Fig. 16-1 is to monitor the wave forms (shapes of
the waves) as they move to the right. Alternatively, we could monitor the motion of
an element of the string as the element oscillates up and down while a wave passes
through it. We would find that the displacement of every such oscillating string ele-
ment is perpendicular to the direction of travel of the wave, as indicated in Fig. 16-1b.
This motion is said to be transverse, and the wave is said to be a transverse wave.
Longitudinal Waves. Figure 16-2 shows how a sound wave can be produced
by a piston in a long, air-filled pipe. If you suddenly move the piston rightward
and then leftward, you can send a pulse of sound along the pipe. The rightward
motion of the piston moves the elements of air next to it rightward, changing the
air pressure there. The increased air pressure then pushes rightward on the
elements of air somewhat farther along the pipe. Moving the piston leftward
reduces the air pressure next to it. As a result, first the elements nearest the
piston and then farther elements move leftward. Thus, the motion of the air and
the change in air pressure travel rightward along the pipe as a pulse.
If you push and pull on the piston in simple harmonic motion, as is being
done in Fig. 16-2, a sinusoidal wave travels along the pipe. Because the motion of
the elements of air is parallel to the direction of the wave’s travel, the motion
is said to be longitudinal, and the wave is said to be a longitudinal wave. In this
chapter we focus on transverse waves, and string waves in particular; in
Chapter 17 we focus on longitudinal waves, and sound waves in particular.
Both a transverse wave and a longitudinal wave are said to be traveling
waves because they both travel from one point to another, as from one end of the
string to the other end in Fig. 16-1 and from one end of the pipe to the other end
in Fig. 16-2. Note that it is the wave that moves from end to end, not the material
(string or air) through which the wave moves.
Wavelength and Frequency
To completely describe a wave on a string (and the motion of any element along
its length), we need a function that gives the shape of the wave. This means that
we need a relation in the form
yh(x,t), (16-1)
in which yis the transverse displacement of any string element as a function hof
the time tand the position xof the element along the string. In general, a sinu-
soidal shape like the wave in Fig.16-1bcan be described with hbeing either a sine
or cosine function; both give the same general shape for the wave. In this chapter
we use the sine function.
Sinusoidal Function. Imagine a sinusoidal wave like that of Fig. 16-1btraveling
in the positive direction of an xaxis. As the wave sweeps through succeeding ele-
ments (that is, very short sections) of the string, the elements oscillate parallel to the y
axis.At time t, the displacement yof the element located at position xis given by
y(x,t)ymsin(kx vt). (16-2)
Because this equation is written in terms of position x, it can be used to find the
displacements of all the elements of the string as a function of time. Thus, it can
tell us the shape of the wave at any given time.
v
:
Figure 16-2 A sound wave is set up in an air-
filled pipe by moving a piston back and
forth. Because the oscillations of an ele-
ment of the air (represented by the dot) are
parallel to the direction in which the wave
travels,the wave is a longitudinal wave.
Air
v
447
16-1 TRANSVERSE WAVES
The names of the quantities in Eq. 16-2 are displayed in Fig. 16-3 and defined
next. Before we discuss them, however, let us examine Fig. 16-4, which shows five
“snapshots” of a sinusoidal wave traveling in the positive direction of an xaxis.
The movement of the wave is indicated by the rightward progress of the short
arrow pointing to a high point of the wave. From snapshot to snapshot, the short
arrow moves to the right with the wave shape, but the string moves only parallel
to the yaxis.To see that, let us follow the motion of the red-dyed string element at
x0. In the first snapshot (Fig. 16-4a), this element is at displacement y0.
In the next snapshot, it is at its extreme downward displacement because a valley
(or extreme low point) of the wave is passing through it. It then moves back up
through y0. In the fourth snapshot, it is at its extreme upward displacement
because a peak (or extreme high point) of the wave is passing through it. In the
fifth snapshot, it is again at y0, having completed one full oscillation.
Amplitude and Phase
The amplitude ymof a wave, such as that in Fig. 16-4 , is the magnitude of the
maximum displacement of the elements from their equilibrium positions as the
wave passes through them. (The subscript mstands for maximum.) Because ymis
a magnitude, it is always a positive quantity, even if it is measured downward
instead of upward as drawn in Fig.16-4a.
The phase of the wave is the argument kx vtof the sine in Eq. 16-2.As the
wave sweeps through a string element at a particular position x, the phase
changes linearly with time t. This means that the sine also changes, oscillating
between 1 and 1. Its extreme positive value (1) corresponds to a peak of the
wave moving through the element; at that instant the value of yat position xis ym.
Its extreme negative value (1) corresponds to a valley of the wave moving
through the element; at that instant the value of yat position xis ym. Thus, the
sine function and the time-dependent phase of a wave correspond to the oscilla-
tion of a string element, and the amplitude of the wave determines the extremes
of the element’s displacement.
Caution: When evaluating the phase, rounding off the numbers before you
evaluate the sine function can throw of the calculation considerably.
Wavelength and Angular Wave Number
The wavelength lof a wave is the distance (parallel to the direction of the wave’s
travel) between repetitions of the shape of the wave (or wave shape). A typical
wavelength is marked in Fig. 16-4a, which is a snapshot of the wave at time t0.
At that time, Eq. 16-2 gives, for the description of the wave shape,
y(x,0)ymsin kx. (16-3)
By definition, the displacement yis the same at both ends of this wave-
length—that is, at xx1and xx1l.Thus,by Eq. 16-3,
ymsin kx1ymsin k(x1l)
ymsin(kx1kl). (16-4)
A sine function begins to repeat itself when its angle (or argument) is increased
by 2prad, so in Eq. 16-4 we must have kl2p,or
(angular wave number). (16-5)
We call kthe angular wave number of the wave; its SI unit is the radian per meter,
or the inverse meter. (Note that the symbol khere does not represent a spring
constant as previously.)
Notice that the wave in Fig. 16-4 moves to the right by lfrom one snapshot
to the next.Thus, by the fifth snapshot, it has moved to the right by 1l.
1
4
k2p
l
Figure 16-4 Five “snapshots” of a string wave
traveling in the positive direction of an
xaxis.The amplitude ymis indicated.A
typical wavelength l, measured from an
arbitrary position x1, is also indicated.
x
y
ymx1
λ
(a)
x
y
(b)
x
y
(c)
x
y
(d)
x
y
(
e
)
Watch this spot in this
series of snapshots.
Figure 16-3 The names of the quantities in
Eq. 16-2, for a transverse sinusoidal wave.
Displacement
Amplitude
Angular
wave number
Position
Time
Angular
fre
q
uency
Phase
Oscillating
term
y(x,t) = ymsin (kx – t)
ω
ω
448 CHAPTER 16 WAVES—I
Period, Angular Frequency, and Frequency
Figure 16-5 shows a graph of the displacement yof Eq. 16-2 versus time tat a
certain position along the string, taken to be x0. If you were to monitor the
string, you would see that the single element of the string at that position moves
up and down in simple harmonic motion given by Eq. 16-2 with x0:
y(0, t)ymsin(vt)
ymsin vt(x0). (16-6)
Here we have made use of the fact that sin(a)sin a, where ais any angle.
Figure 16-5 is a graph of this equation, with displacement plotted versus time; it
does not show the shape of the wave. (Figure 16-4 shows the shape and is a
picture of reality; Fig. 16-5 is a graph and thus an abstraction.)
We define the period of oscillation Tof a wave to be the time any string
element takes to move through one full oscillation.A typical period is marked on
the graph of Fig. 16-5. Applying Eq. 16-6 to both ends of this time interval and
equating the results yield
ymsin vt1ymsin v(t1T)
ymsin(vt1vT). (16-7)
This can be true only if vT2p, or if
(angular frequency). (16-8)
We call vthe angular frequency of the wave; its SI unit is the radian per second.
Look back at the five snapshots of a traveling wave in Fig. 16-4. The time
between snapshots is T. Thus, by the fifth snapshot, every string element has
made one full oscillation.
The frequency fof a wave is defined as 1/Tand is related to the angular
frequency vby
(frequency). (16-9)
Like the frequency of simple harmonic motion in Chapter 15, this frequency fis a
number of oscillations per unit time—here,the number made by a string element
as the wave moves through it. As in Chapter 15, fis usually measured in hertz or
its multiples, such as kilohertz.
f1
Tv
2p
1
4
v2p
T
Checkpoint 1
The figure is a composite of three snapshots, each of
a wave traveling along a particular string.The
phases for the waves are given by (a) 2x4t,
(b) 4x8t, and (c) 8x16t.Which phase
corresponds to which wave in the figure?
1 2 3
x
y
Figure 16-5 A graph of the displacement of
the string element at x0 as a function of
time, as the sinusoidal wave of Fig. 16-4
passes through the element.The amplitude
ymis indicated.A typical period T, mea-
sured from an arbitrary time t1, is also
indicated.
t
y
t
1
y
m
T
This is a graph,
not a snapshot.
Phase Constant
When a sinusoidal traveling wave is given by the wave function of Eq. 16-2, the
wave near x0 looks like Fig. 16-6awhen t0. Note that at x0, the displace-
ment is y0 and the slope is at its maximum positive value. We can generalize
Eq. 16-2 by inserting a phase constant fin the wave function:
yymsin(kx vtf). (16-10)
Figure 16-6 A sinusoidal traveling wave at
t0 with a phase constant fof (a) 0 and
(b)p/5 rad.
x
x
y
y
(a)
(b)
The effect of the
phase constant
is to shift the wave.
φ
449
16-1 TRANSVERSE WAVES
The value of fcan be chosen so that the function gives some other displacement
and slope at x0 when t0. For example, a choice of fp/5 rad gives the
displacement and slope shown in Fig. 16-6bwhen t0. The wave is still
sinusoidal with the same values of ym,k, and v, but it is now shifted from what
you see in Fig. 16-6a(where f0). Note also the direction of the shift.A positive
value of fshifts the curve in the negative direction of the xaxis; a negative value
shifts the curve in the positive direction.
The Speed of a Traveling Wave
Figure 16-7 shows two snapshots of the wave of Eq. 16-2, taken a small time
interval tapart. The wave is traveling in the positive direction of x(to the right
in Fig. 16-7), the entire wave pattern moving a distance xin that direction
during the interval t. The ratio x/t(or, in the differential limit, dx/dt) is the
wave speed v. How can we find its value?
As the wave in Fig.16-7 moves, each point of the moving wave form, such as point
Amarked on a peak,retains its displacement y.(Points on the string do not retain their
displacement, but points on the wave form do.) If point Aretains its displacement as it
moves, the phase in Eq.16-2 giving it that displacement must remain a constant:
kx vta constant. (16-11)
Note that although this argument is constant, both xand tare changing. In fact,
as tincreases, xmust also, to keep the argument constant. This confirms that the
wave pattern is moving in the positive direction of x.
To find the wave speed v, we take the derivative of Eq. 16-11, getting
or (16-12)
Using Eq. 16-5 (k2p/l) and Eq. 16-8 (v2p/T), we can rewrite the wave
speed as
(wave speed). (16-13)
The equation vl/Ttells us that the wave speed is one wavelength per period;
the wave moves a distance of one wavelength in one period of oscillation.
Equation 16-2 describes a wave moving in the positive direction of x.We can
find the equation of a wave traveling in the opposite direction by replacing tin
Eq. 16-2 with t.This corresponds to the condition
kx vta constant, (16-14)
which (compare Eq. 16-11) requires that x decrease with time. Thus, a wave trav-
eling in the negative direction of xis described by the equation
y(x,t)ymsin(kx vt). (16-15)
If you analyze the wave of Eq. 16-15 as we have just done for the wave of
Eq. 16-2, you will find for its velocity
(16-16)
The minus sign (compare Eq. 16-12) verifies that the wave is indeed moving in the
negative direction of xand justifies our switching the sign of the time variable.
dx
dt v
k.
vv
kl
Tlf
dx
dt vv
k.
kdx
dt v0
Figure 16-7 Two snapshots of the wave of
Fig. 16-4, at time t0 and then at time
tt.As the wave moves to the right at
velocity , the entire curve shifts a distance
xduring t. Point A“rides” with the wave
form, but the string elements move only up
and down.
v
:
x
yΔ x
A
Wave at t = 0
Wave at t = Δt
v
450 CHAPTER 16 WAVES—I
Consider now a wave of arbitrary shape, given by
y(x,t)h(kx vt), (16-17)
where hrepresents any function, the sine function being one possibility. Our
previous analysis shows that all waves in which the variables xand tenter
into the combination kx vtare traveling waves. Furthermore, all traveling
waves must be of the form of Eq. 16-17. Thus, y(x,t)represents a
possible (though perhaps physically a little bizarre) traveling wave. The function
y(x,t)sin(ax2bt), on the other hand, does not represent a traveling wave.
1ax bt
Checkpoint 2
Here are the equations of three waves:
(1) y(x,t)2 sin(4x2t), (2) y(x,t)sin(3x4t), (3) y(x,t)2 sin(3x3t).
Rank the waves according to their (a) wave speed and (b) maximum speed perpendi-
cular to the wave’s direction of travel (the transverse speed), greatest first.
straction, showing us motion spread out over time. From it
we can determine the period Tof the string element in its
SHM and thus also of the wave itself. From Twe can then
find angular frequency v(2p/T) in Eq. 16-18. (3) The
phase constant fis set by the displacement of the string at
x0 and t0.
Amplitude: From either Fig. 16-8aor 16-8bwe see that the
maximum displacement is 3.0 mm. Thus, the wave’s ampli-
tude xm3.0 mm.
Wavelength: In Fig. 16-8a, the wavelength lis the distance
along the xaxis between repetitions in the pattern.The easi-
est way to measure lis to find the distance from one cross-
ing point to the next crossing point where the string has the
same slope. Visually we can roughly measure that distance
with the scale on the axis. Instead, we can lay the edge of a
Sample Problem 16.01 Determining the quantities in an equation for a transverse wave
A transverse wave traveling along an xaxis has the form
given by
yymsin(kx vtf). (16-18)
Figure 16-8agives the displacements of string elements as a
function of x, all at time t0. Figure 16-8bgives the
displacements of the element at x0 as a function of t. Find
the values of the quantities shown in Eq. 16-18, including the
correct choice of sign.
KEY IDEAS
(1) Figure 16-8ais effectively a snapshot of reality (some-
thing that we can see), showing us motion spread out over
the xaxis. From it we can determine the wavelength lof the
wave along that axis, and then we can find the angular wave
number k(2p/l) in Eq. 16-18. (2) Figure 16-8bis an ab-
Figure 16-8 (a) A snapshot of the displacement yversus position xalong a string, at time t0. (b) A graph of displacement yversus time
tfor the string element at x0.
(a)
–8
2
–2
0
y(mm)
x(mm)
–4 48
(b)
0
y(mm)
t(ms)
–20 10 20–10
–3
–2
2
451
16-1 TRANSVERSE WAVES
crease (mentally slide the curve slightly rightward). If,
instead, the wave is moving leftward, then just after the snap-
shot, the depth at x0 should decrease. Now let’s check the
graph in Fig. 16-8b. It tells us that just after t0, the depth in-
creases.Thus, the wave is moving rightward, in the positive di-
rection of x,and we choose the minus sign in Eq. 16-18.
Phase constant: The value of fis set by the conditions at
x0 at the instant t0. From either figure we see that at
that location and time, y2.0 mm. Substituting these
three values and also ym3.0 mm into Eq. 16-18 gives us
2.0 mm (3.0 mm) sin(0 0f)
or fsin10.73 rad.
Note that this is consistent with the rule that on a plot of y
versus x, a negative phase constant shifts the normal sine
function rightward, which is what we see in Fig. 16-8a.
Equation: Now we can fill out Eq. 16-18:
y(3.0 mm) sin(200pxl00pt0.73 rad), (Answer)
with xin meters and tin seconds.
(2
3)
paper sheet on the graph, mark those crossing points, slide
the sheet to align the left-hand mark with the origin, and
then read off the location of the right-hand mark. Either
way we find l10 mm. From Eq.16-5, we then have
Period: The period Tis the time interval that a string ele-
ment’s SHM takes to begin repeating itself. In Fig. 16-8b,T
is the distance along the taxis from one crossing point to the
next crossing point where the plot has the same slope.
Measuring the distance visually or with the aid of a sheet of
paper,we find T20 ms. From Eq. 16-8, we then have
Direction of travel: To find the direction, we apply a bit of
reasoning to the figures. In the snapshot at t0 given in
Fig. 16-8a, note that if the wave is moving rightward, then just
after the snapshot, the depth of the wave at x0 should in-
v2p
T2p
0.020 s 100p rad/s.
k2p
l2p
0.010 m 200p rad/m.
Next, substituting numerical values but suppressing the
units, which are SI, we write
u(2.72)(0.00327) cos[(72.1)(0.225) (2.72)(18.9)]
0.00720 m/s 7.20 mm/s. (Answer)
Thus, at t18.9 s our string element is moving in the
positive direction of ywith a speed of 7.20 mm/s.
(Caution: In evaluating the cosine function, we keep all the
significant figures in the argument or the calculation can be
off considerably. For example, round off the numbers to two
significant figures and then see what you get for u.)
(b) What is the transverse acceleration ayof our string
element at t18.9 s?
KEY IDEA
The transverse acceleration ayis the rate at which the ele-
ment’s transverse velocity is changing.
Calculations: From Eq. 16-20, again treating xas a constant
but allowing tto vary, we find
(16-21)
Substituting numerical values but suppressing the units,
which are SI, we have
ay5 2(2.72)2(0.00327) sin[(72.1)(0.225) (2.72)(18.9)]
0.0142 m/s214.2 mm/s2. (Answer)
ayu
tv2ym sin (kx vt).
Sample Problem 16.02 Transverse velocity and transverse acceleration of a string element
A wave traveling along a string is described by
y(x,t) = (0.00327 m) sin(72.1x2.72t),
in which the numerical constants are in SI units (72.1 rad/m
and 2.72 rad/s).
(a) What is the transverse velocity uof the string element
at x22.5 cm at time t18.9 s? (This velocity, which is
associated with the transverse oscillation of a string
element, is parallel to the yaxis. Don’t confuse it with v,
the constant velocity at which the wave form moves along
the xaxis.)
KEY IDEAS
The transverse velocity uis the rate at which the
displacement yof the element is changing. In general, that
displacement is given by
y(x,t) = ymsin(kx vt). (16-19)
For an element at a certain location x, we find the rate of
change of yby taking the derivative of Eq. 16-19 with re-
spect to twhile treating xas a constant. A derivative taken
while one (or more) of the variables is treated as a constant
is called a partial derivative and is represented by a symbol
such as rather than d/dt.
Calculations: Here we have
(16-20)uy
tvym cos(kx vt).
/t
452 CHAPTER 16 WAVES—I
16-2 WAVE SPEED ON A STRETCHED STRING
After reading this module, you should be able to . . .
16.14 Calculate the linear density mof a uniform string in
terms of the total mass and total length.
l6.15 Apply the relationship between wave speed v, tension t,
and linear density m.
●The speed of a wave on a stretched string is set by
properties of the string, not properties of the wave such as
frequency or amplitude.
●The speed of a wave on a string with tension tand linear
density mis
vAt
m.
Learning Objectives
Key Ideas
Wave Speed on a Stretched String
The speed of a wave is related to the wave’s wavelength and frequency by Eq.
16-13, but it is set by the properties of the medium. If a wave is to travel through
a medium such as water, air, steel, or a stretched string, it must cause the particles
of that medium to oscillate as it passes, which requires both mass (for kinetic en-
ergy) and elasticity (for potential energy).Thus, the mass and elasticity determine
how fast the wave can travel. Here,we find that dependency in two ways.
Dimensional Analysis
In dimensional analysis we carefully examine the dimensions of all the physical
quantities that enter into a given situation to determine the quantities they pro-
duce. In this case, we examine mass and elasticity to find a speed v, which has the
dimension of length divided by time, or LT 1.
For the mass, we use the mass of a string element, which is the mass mof the
string divided by the length lof the string.We call this ratio the linear density mof
the string.Thus, mm/l, its dimension being mass divided by length, ML1.
You cannot send a wave along a string unless the string is under tension,
which means that it has been stretched and pulled taut by forces at its two ends.
The tension tin the string is equal to the common magnitude of those two forces.
As a wave travels along the string, it displaces elements of the string by causing
additional stretching, with adjacent sections of string pulling on each other
because of the tension. Thus, we can associate the tension in the string with the
stretching (elasticity) of the string. The tension and the stretching forces it pro-
duces have the dimension of a force—namely, MLT 2(from Fma).
We need to combine m(dimension ML1) and t(dimension MLT 2) to get v
(dimension LT 1).A little juggling of various combinations suggests
(16-22)
in which Cis a dimensionless constant that cannot be determined with dimen-
sional analysis. In our second approach to determining wave speed, you will see
that Eq. 16-22 is indeed correct and that C1.
vCAt
m,
Additional examples, video, and practice available at WileyPLUS
From part (a) we learn that at t18.9 s our string element is
moving in the positive direction of y, and here we learn that
it is slowing because its acceleration is in the opposite
direction of u.
453
16-2 WAVE SPEED ON A STRETCHED STRING
Derivation from Newton’s Second Law
Instead of the sinusoidal wave of Fig. 16-1b, let us consider a single symmetrical
pulse such as that of Fig. 16-9, moving from left to right along a string with
speed v. For convenience, we choose a reference frame in which the pulse
remains stationary; that is, we run along with the pulse, keeping it constantly
in view. In this frame, the string appears to move past us, from right to left in
Fig.16-9, with speed v.
Consider a small string element of length lwithin the pulse, an element that
forms an arc of a circle of radius Rand subtending an angle 2uat the center of
that circle. A force with a magnitude equal to the tension in the string pulls
tangentially on this element at each end. The horizontal components of these
forces cancel, but the vertical components add to form a radial restoring force .
In magnitude,
(force), (16-23)
where we have approximated sin uas ufor the small angles uin Fig. 16-9. From
that figure, we have also used 2ul/R.The mass of the element is given by
mml(mass), (16-24)
where mis the string’s linear density.
At the moment shown in Fig. 16-9, the string element lis moving in an arc of a
circle.Thus, it has a centripetal acceleration toward the center of that circle, given by
(acceleration). (16-25)
Equations 16-23, 16-24, and 16-25 contain the elements of Newton’s second
law. Combining them in the form
force mass acceleration
gives
Solving this equation for the speed vyields
(speed), (16-26)
in exact agreement with Eq. 16-22 if the constant Cin that equation is given the
value unity. Equation 16-26 gives the speed of the pulse in Fig. 16-9 and the speed
of any other wave on the same string under the same tension.
Equation 16-26 tells us:
vAt
m
t
l
R(m
l)v2
R.
av2
R
F2(t
sin u)t(2u)tl
R
F
:
:
Figure 16-9 A symmetrical pulse,viewed
from a reference frame in which the pulse
is stationary and the string appears to move
right to left with speed v.We find speed v
by applying Newton’s second law to a
string element of length l, located at the
top of the pulse.
τ
θ
R
lΔ
O
τ
θ
v
The speed of a wave along a stretched ideal string depends only on the tension
and linear density of the string and not on the frequency of the wave.
Checkpoint 3
You send a traveling wave along a particular string by oscillating one end. If you
increase the frequency of the oscillations, do (a) the speed of the wave and (b) the
wavelength of the wave increase, decrease, or remain the same? If, instead, you
increase the tension in the string,do (c) the speed of the wave and (d) the wavelength
of the wave increase, decrease, or remain the same?
The frequency of the wave is fixed entirely by whatever generates the wave (for
example, the person in Fig. 16-1b). The wavelength of the wave is then fixed by
Eq. 16-13 in the form lv/f.
454 CHAPTER 16 WAVES—I
Energy and Power of a Wave Traveling Along a String
When we set up a wave on a stretched string, we provide energy for the motion of
the string. As the wave moves away from us, it transports that energy as both
kinetic energy and elastic potential energy. Let us consider each form in turn.
Kinetic Energy
A string element of mass dm, oscillating transversely in simple harmonic motion
as the wave passes through it, has kinetic energy associated with its transverse
velocity . When the element is rushing through its y0 position (element bin
Fig. 16-10), its transverse velocity—and thus its kinetic energy—is a maximum.
When the element is at its extreme position yym(as is element a), its trans-
verse velocity—and thus its kinetic energy—is zero.
Elastic Potential Energy
To send a sinusoidal wave along a previously straight string, the wave must neces-
sarily stretch the string.As a string element of length dx oscillates transversely, its
length must increase and decrease in a periodic way if the string element is to fit
the sinusoidal wave form. Elastic potential energy is associatzed with these
length changes, just as for a spring.
When the string element is at its yymposition (element ain Fig. 16-10), its
length has its normal undisturbed value dx, so its elastic potential energy is zero.
However, when the element is rushing through its y0 position, it has maximum
stretch and thus maximum elastic potential energy.
Energy Transport
The oscillating string element thus has both its maximum kinetic energy and its
maximum elastic potential energy at y0. In the snapshot of Fig. 16-10, the
regions of the string at maximum displacement have no energy, and the regions at
zero displacement have maximum energy. As the wave travels along the string,
forces due to the tension in the string continuously do work to transfer energy
from regions with energy to regions with no energy.
As in Fig. 16-1b, let’s set up a wave on a string stretched along a horizontal x
axis such that Eq. 16-2 applies. As we oscillate one end of the string, we continu-
ously provide energy for the motion and stretching of the string—as the string
sections oscillate perpendicularly to the xaxis, they have kinetic energy and elas-
tic potential energy.As the wave moves into sections that were previously at rest,
energy is transferred into those new sections. Thus, we say that the wave trans-
ports the energy along the string.
The Rate of Energy Transmission
The kinetic energy dK associated with a string element of mass dm is given by
dK dm u2, (16-27)
1
2
u
:
16-3 ENERGY AND POWER OF A WAVE TRAVELING ALONG A STRING
After reading this module, you should be able to . . .
16.16 Calculate the average rate at which energy is transported by a transverse wave.
●The average power of, or average rate at which energy is
transmitted by, a sinusoidal wave on a stretched string is
Learning Objective
Key Idea
Figure 16-10 A snapshot of a traveling wave
on a string at time t0. String element ais
at displacement yym, and string element
bis at displacement y0.The kinetic en-
ergy of the string element at each position
depends on the transverse velocity of the
element.The potential energy depends on
the amount by which the string element is
stretched as the wave passes through it.
y
y
m
0
dx
b
dx
a
λ
x
v
given by
Pavg 1
2mvv2y2
m.
455
16-3 ENERGY AND POWER OF A WAVE TRAVELING ALONG A STRING
where uis the transverse speed of the oscillating string element. To find u,we
differentiate Eq. 16-2 with respect to time while holding xconstant:
(16-28)
Using this relation and putting dm mdx, we rewrite Eq. 16-27 as
dK (mdx)(vym)2cos2(kx vt). (16-29)
Dividing Eq. 16-29 by dt gives the rate at which kinetic energy passes through
a string element, and thus the rate at which kinetic energy is carried along by the
wave.The dx/dt that then appears on the right of Eq. 16-29 is the wave speed v, so
(16-30)
The average rate at which kinetic energy is transported is
mvv2y2
m. (16-31)
Here we have taken the average over an integer number of wavelengths and
have used the fact that the average value of the square of a cosine function over
an integer number of periods is .
Elastic potential energy is also carried along with the wave, and at the same
average rate given by Eq. 16-31. Although we shall not examine the proof, you
should recall that, in an oscillating system such as a pendulum or a spring–block
system, the average kinetic energy and the average potential energy are equal.
The average power, which is the average rate at which energy of both kinds
is transmitted by the wave,is then
(16-32)
or,from Eq. 16-31,
Pavg mvv2y2
m(average power). (16-33)
The factors mand vin this equation depend on the material and tension of the
string.The factors vand ymdepend on the process that generates the wave.The de-
pendence of the average power of a wave on the square of its amplitude and also on
the square of its angular frequency is a general result,true for waves of all types.
1
2
Pavg 2
dK
dt
avg
1
2
1
4
dK
dt
avg
1
2mvv2y2
m [cos2(kx vt)]avg
dK
dt 1
2
v
2y2
m cos2(kx
t).
1
2
u≠y
≠tvym cos(kx vt).
angular frequency vand wave speed v.From Eq. 16-9,
v2pf(2p)(120 Hz) 754 rad/s.
From Eq. 16-26 we have
Equation 16-33 then yields
Pavg mvv2y2
m
( )(0.525 kg/m)(9.26 m/s)(754 rad/s)2(0.0085 m)2
100 W. (Answer)
1
2
1
2
vA
A45 N
0.525 kg/m 9.26 m/s.
Sample Problem 16.03 Average power of a transverse wave
A string has linear density m525 g/m and is under tension
t45 N. We send a sinusoidal wave with frequency f120 Hz
and amplitude ym8.5 mm along the string. At what average
rate does the wave transport energy?
KEY IDEA
The average rate of energy transport is the average power
Pavg as given by Eq. 16-33.
Calculations: To use Eq. 16-33, we first must calculate
Additional examples, video, and practice available at WileyPLUS
The Wave Equation
As a wave passes through any element on a stretched string, the element moves
perpendicularly to the wave’s direction of travel (we are dealing with a trans-
verse wave). By applying Newton’s second law to the element’s motion, we can
derive a general differential equation, called the wave equation, that governs the
travel of waves of any type.
Figure 16-11ashows a snapshot of a string element of mass dm and length
as a wave travels along a string of linear density mthat is stretched along a hori-
zontal xaxis. Let us assume that the wave amplitude is small so that the element
can be tilted only slightly from the xaxis as the wave passes. The force 2on the
right end of the element has a magnitude equal to tension tin the string and is
directed slightly upward. The force 1on the left end of the element also has
a magnitude equal to the tension tbut is directed slightly downward. Because of
the slight curvature of the element, these two forces are not simply in opposite di-
rection so that they cancel. Instead, they combine to produce a net force that
causes the element to have an upward acceleration ay. Newton’s second law writ-
ten for ycomponents (Fnet,ymay) gives us
F2yF1ydm ay. (16-34)
Let’s analyze this equation in parts, first the mass dm, then the acceleration com-
ponent ay, then the individual force components F2yand F1y, and then finally the
net force that is on the left side of Eq. 16-34.
Mass. The element’s mass dm can be written in terms of the string’s linear
density mand the element’s length as dm m. Because the element can have
only a slight tilt, dx (Fig. 16-11a) and we have the approximation
dm mdx. (16-35)
F
:
F
:
456 CHAPTER 16 WAVES—I
16-4 THE WAVE EQUATION
After reading this module, you should be able to . . .
16.17 For the equation giving a string-element displacement
as a function of position xand time t, apply the relationship
between the second derivative with respect to xand the
second derivative with respect to t.
●The general differential equation that governs the travel of waves
of all types is 2y
x21
v2
2y
t2.
Here the waves travel along an xaxis and oscillate parallel to
the yaxis, and they move with speed v, in either the positive x
direction or the negative xdirection.
Learning Objective
Key Idea
Figure 16-11 (a) A string element as a sinusoidal transverse wave travels on a stretched string.
Forces 1and 2act at the left and right ends, producing acceleration having a vertical
component ay.(b) The force at the element’s right end is directed along a tangent to the ele-
ment’s right side.
a
:
F
:
F
:
y
x
dx
F1
F2
ay
(a)
y
x
F2
F2x
F2y
(b)
Tangent line
457
16-4 THE WAVE EQUATION
Acceleration. The acceleration ayin Eq. 16-34 is the second derivative of the
displacement ywith respect to time:
(16-36)
Forces. Figure 16-11bshows that 2is tangent to the string at the right end
of the string element.Thus we can relate the components of the force to the string
slope S2at the right end as
(16-37)
We can also relate the components to the magnitude F2(t) with
or (16-38)
However,because we assume that the element is only slightly tilted, F2yF2xand
therefore we can rewrite Eq. 16-38 as
tF2x. (16-39)
Substituting this into Eq. 16-37 and solving for F2yyield
F2ytS2. (16-40)
Similar analysis at the left end of the string element gives us
F1ytS1. (16-41)
Net Force. We can now substitute Eqs. 16-35, 16-36, 16-40, and 16-41 into
Eq. 16-34 to write
or (16-42)
Because the string element is short, slopes S2and S1differ by only a differential
amount dS, where Sis the slope at any point:
(16-43)
First replacing S2S1in Eq. 16-42 with dS and then using Eq. 16-43 to substitute
dy/dx for S, we find
and . (16-44)
In the last step, we switched to the notation of partial derivatives because on
the left we differentiate only with respect to xand on the right we differenti-
ate only with respect to t. Finally, substituting from Eq. 16-26 (v), we
find
(wave equation). (16-45)
This is the general differential equation that governs the travel of waves of all
types.
≠2y
≠x21
v2
≠2y
≠t2
1t/m
≠2y
≠x2m
t
≠2y
≠t2
d(dy/dx)
dx m
t
d2y
dt2,
dS
dx m
t
d2y
dt2,
Sdy
dx .
S2S1
dx m
t
d2y
dt2.
tS2tS1(m
dx)d2y
dt2,
t2F2
2xF2
2y.
F22F2
2xF2
2y
F2y
F2x
S2.
F
:
ayd2y
dt2.
458 CHAPTER 16 WAVES—I
The Principle of Superposition for Waves
It often happens that two or more waves pass simultaneously through the same
region.When we listen to a concert, for example, sound waves from many instru-
ments fall simultaneously on our eardrums. The electrons in the antennas of our
radio and television receivers are set in motion by the net effect of many electro-
magnetic waves from many different broadcasting centers. The water of a lake or
harbor may be churned up by waves in the wakes of many boats.
Suppose that two waves travel simultaneously along the same stretched
string. Let y1(x,t) and y2(x,t) be the displacements that the string would
experience if each wave traveled alone. The displacement of the string when the
waves overlap is then the algebraic sum
y(x,t)y1(x,t)y2(x,t). (16-46)
This summation of displacements along the string means that
This is another example of the principle of superposition, which says that when
several effects occur simultaneously, their net effect is the sum of the individual
effects. (We should be thankful that only a simple sum is needed. If two effects
somehow amplified each other, the resulting nonlinear world would be very diffi-
cult to manage and understand.)
Figure 16-12 shows a sequence of snapshots of two pulses traveling in
opposite directions on the same stretched string. When the pulses overlap, the
resultant pulse is their sum. Moreover,
Overlapping waves algebraically add to produce a resultant wave (or net wave).
Overlapping waves do not in any way alter the travel of each other.
Figure 16-12 A series of snapshots that
show two pulses traveling in opposite
directions along a stretched string. The
superposition principle applies as the
pulses move through each other.
When two waves overlap,
we see the resultant wave,
not the individual waves.
●When two or more waves traverse the same medium, the
displacement of any particle of the medium is the sum of the
displacements that the individual waves would give it, an
effect known as the principle of superposition for waves.
●Two sinusoidal waves on the same string exhibit
interference, adding or canceling according to the
principle of superposition. If the two are traveling in the
same direction and have the same amplitude ymand
frequency (hence the same wavelength) but differ in phase by
a phase constant f, the result is a single wave with this same
frequency:
y(x,t)[2ymcos f] sin(kx vtf).
If f0, the waves are exactly in phase and their interference
is fully constructive; if fprad, they are exactly out of phase
and their interference is fully destructive.
1
2
1
2
Key Ideas
16-5 INTERFERENCE OF WAVES
After reading this module, you should be able to . . .
16.18 Apply the principle of superposition to show that two
overlapping waves add algebraically to give a resultant
(or net) wave.
16.19 For two transverse waves with the same amplitude and
wavelength and that travel together, find the displacement equa-
tion for the resultant wave and calculate the amplitude in terms
of the individual wave amplitude and the phase difference.
16.20 Describe how the phase difference between two
transverse waves (with the same amplitude and wavelength)
can result in fully constructive interference, fully destructive in-
terference, and intermediate interference.
16.21 With the phase difference between two interfering
waves expressed in terms of wavelengths, quickly
determine the type of interference the waves have.
Learning Objectives
459
16-5 INTERFERENCE OF WAVES
Interference of Waves
Suppose we send two sinusoidal waves of the same wavelength and amplitude in
the same direction along a stretched string. The superposition principle applies.
What resultant wave does it predict for the string?
The resultant wave depends on the extent to which the waves are in phase
(in step) with respect to each other—that is, how much one wave form is
shifted from the other wave form. If the waves are exactly in phase (so that the
peaks and valleys of one are exactly aligned with those of the other), they com-
bine to double the displacement of either wave acting alone. If they are exactly
out of phase (the peaks of one are exactly aligned with the valleys of the other),
they combine to cancel everywhere, and the string remains straight.We call this
phenomenon of combining waves interference, and the waves are said to
interfere. (These terms refer only to the wave displacements; the travel of the
waves is unaffected.)
Let one wave traveling along a stretched string be given by
y1(x,t)ymsin(kx vt) (16-47)
and another,shifted from the first, by
y2(x,t)ymsin(kx vtf). (16-48)
These waves have the same angular frequency v(and thus the same frequency
f), the same angular wave number k(and thus the same wavelength l), and the
same amplitude ym.They both travel in the positive direction of the xaxis, with
the same speed, given by Eq. 16-26. They differ only by a constant angle f, the
phase constant.These waves are said to be out of phase by for to have a phase
difference of f, or one wave is said to be phase-shifted from the other by f.
From the principle of superposition (Eq. 16-46), the resultant wave is the
algebraic sum of the two interfering waves and has displacement
y(x,t)y1(x,t)y2(x,t)
ymsin(kx vt)ymsin(kx vtf). (16-49)
In Appendix E we see that we can write the sum of the sines of two angles aand bas
sin asin b2 sin (ab) cos (ab). (16-50)
Applying this relation to Eq. 16-49 leads to
y(x,t)[2ymcos f] sin(kx vtf). (16-51)
As Fig. 16-13 shows, the resultant wave is also a sinusoidal wave traveling in the
direction of increasing x. It is the only wave you would actually see on the string
(you would not see the two interfering waves of Eqs. 16-47 and 16-48).
1
2
1
2
1
2
1
2
If two sinusoidal waves of the same amplitude and wavelength travel in the same
direction along a stretched string, they interfere to produce a resultant sinusoidal
wave traveling in that direction.
Figure 16-13 The resultant wave of
Eq. 16-51, due to the interference of two
sinusoidal transverse waves, is also a
sinusoidal transverse wave, with an
amplitude and an oscillating term.
Displacement
Magnitude
gives
am
p
litude
y'(x,t) = [2ymcos ] sin(kx – t + )
φ ω φ
Oscillating
term
1
__
2
1
__
2
The resultant wave differs from the interfering waves in two respects: (1) its phase
constant is f, and (2) its amplitude y
mis the magnitude of the quantity in the brack-
ets in Eq.16-51: y
m|2ymcos f|(amplitude). (16-52)
If f0 rad (or 0), the two interfering waves are exactly in phase and Eq.
16-51 reduces to
y(x,t)2ymsin(kx vt)(f0). (16-53)
1
2
1
2
460 CHAPTER 16 WAVES—I
x
y
= 0
y1(x,t)
and
y2(x,t)
φ
(a)
x
y
= rad
y1(x,t)y2(x,t)
φ π
(b)
x
y
= rad
y1(x,t)y2(x,t)
φ π
(c)
2
__
3
x
y
y'(x,t)
(d)
x
y
y'(x,t)
(e)
x
y
y'(x,t)
(f)
Being exactly in phase,
the waves produce a
large resultant wave.
Being exactly out of
phase, they produce
a flat string.
This is an intermediate
situation, with an
intermediate result.
Figure 16-14 Two identical sinusoidal waves,
y1(x,t) and y2(x,t), travel along a string in
the positive direction of an xaxis.They in-
terfere to give a resultant wave y(x,t).
The resultant wave is what is actually
seen on the string.The phase difference f
between the two interfering waves is (a) 0
rad or 0,(b)prad or 180, and (c)prad
or 120.The corresponding result-
ant waves are shown in (d), (e), and (f).
2
3
The two waves are shown in Fig. 16-14a, and the resultant wave is plotted in Fig.
16-14d. Note from both that plot and Eq. 16-53 that the amplitude of the resultant
wave is twice the amplitude of either interfering wave.That is the greatest ampli-
tude the resultant wave can have, because the cosine term in Eqs. 16-51 and 16-52
has its greatest value (unity) when f0. Interference that produces the greatest
possible amplitude is called fully constructive interference.
If fprad (or 180), the interfering waves are exactly out of phase as in Fig.
16-14b.Then cos fbecomes cos p/2 0, and the amplitude of the resultant wave
as given by Eq. 16-52 is zero.We then have, for all values of xand t,
y(x,t)0(fprad). (16-54)
The resultant wave is plotted in Fig. 16-14e. Although we sent two waves along
the string, we see no motion of the string. This type of interference is called fully
destructive interference.
Because a sinusoidal wave repeats its shape every 2prad, a phase difference
of f2prad (or 360) corresponds to a shift of one wave relative to the other
wave by a distance equivalent to one wavelength. Thus, phase differences can be
described in terms of wavelengths as well as angles. For example, in Fig. 16-14b
the waves may be said to be 0.50 wavelength out of phase.Table 16-1 shows some
other examples of phase differences and the interference they produce. Note that
when interference is neither fully constructive nor fully destructive, it is called
intermediate interference. The amplitude of the resultant wave is then interme-
diate between 0 and 2ym. For example, from Table 16-1, if the interfering waves
have a phase difference of 120(fprad 0.33 wavelength), then the result-
ant wave has an amplitude of ym, the same as that of the interfering waves
(see Figs. 16-14cand f).
Two waves with the same wavelength are in phase if their phase
difference is zero or any integer number of wavelengths.Thus, the integer part of
any phase difference expressed in wavelengths may be discarded. For example, a
phase difference of 0.40 wavelength (an intermediate interference, close to fully
destructive interference) is equivalent in every way to one of 2.40 wavelengths,
2
3
1
2
461
and so the simpler of the two numbers can be used in computations. Thus, by
looking at only the decimal number and comparing it to 0, 0.5, or 1.0 wavelength,
you can quickly tell what type of interference two waves have.
16-5 INTERFERENCE OF WAVES
Checkpoint 4
Here are four possible phase differences between two identical waves, expressed in
wavelengths: 0.20, 0.45, 0.60, and 0.80. Rank them according to the amplitude of the
resultant wave, greatest first.
Table 16-1 Phase Difference and Resulting Interference Typesa
Amplitude
Phase Difference, in of Resultant Type of
Degrees Radians Wavelengths Wave Interference
00 0 2ymFully constructive
120 p0.33 ymIntermediate
180 p0.50 0 Fully destructive
240 p0.67 ymIntermediate
360 2p1.00 2ymFully constructive
865 15.1 2.40 0.60ymIntermediate
aThe phase difference is between two otherwise identical waves, with amplitude ym, moving in the
same direction.
4
3
2
3
(b) What phase difference, in radians and wavelengths, will
give the resultant wave an amplitude of 4.9 mm?
Calculations: Now we are given y
mand seek f. From Eq.
16-52,
y
m|2ymcos f|,
we now have
4.9 mm (2)(9.8 mm) cos f,
which gives us (with a calculator in the radian mode)
2.636 rad 2.6 rad. (Answer)
There are two solutions because we can obtain the same re-
sultant wave by letting the first wave lead (travel ahead of)
or lag (travel behind) the second wave by 2.6 rad. In wave-
lengths, the phase difference is
0.42 wavelength. (Answer)
f
2p
rad/wavelength 2.636 rad
2p
rad/wavelength
f2 cos14.9 mm
(2)(9.8 mm)
1
2
1
2
Sample Problem 16.04 Interference of two waves, same direction, same amplitude
Two identical sinusoidal waves, moving in the same
direction along a stretched string, interfere with each other.
The amplitude ymof each wave is 9.8 mm, and the phase
difference fbetween them is 100.
(a) What is the amplitude y
mof the resultant wave due to the
interference,and what is the type of this interference?
KEY IDEA
These are identical sinusoidal waves traveling in the same
direction along a string, so they interfere to produce a
sinusoidal traveling wave.
Calculations: Because they are identical, the waves have
the same amplitude. Thus, the amplitude y
mof the resultant
wave is given by Eq. 16-52:
y
m|2ymcos f||(2)(9.8 mm) cos(100/2)|
13 mm. (Answer)
We can tell that the interference is intermediate in two ways.
The phase difference is between 0 and 180, and, correspond-
ingly, the amplitude y
mis between 0 and 2ym(19.6 mm).
1
2
Additional examples, video, and practice available at WileyPLUS
462 CHAPTER 16 WAVES—I
Phasors
Adding two waves as discussed in the preceding module is strictly limited to
waves with identical amplitudes. If we have such waves, that technique is easy
enough to use, but we need a more general technique that can be applied to any
waves, whether or not they have the same amplitudes. One neat way is to use
phasors to represent the waves. Although this may seem bizarre at first, it is es-
sentially a graphical technique that uses the vector addition rules of Chapter 3 in-
stead of messy trig additions.
Aphasor is a vector that rotates around its tail, which is pivoted at the origin
of a coordinate system. The magnitude of the vector is equal to the amplitude ym
of the wave that it represents.The angular speed of the rotation is equal to the an-
gular frequency vof the wave. For example,the wave
y1(x,t)ym1sin(kx vt) (16-55)
is represented by the phasor shown in Figs. 16-15ato d. The magnitude of the
phasor is the amplitude ym1of the wave. As the phasor rotates around the origin
at angular speed v, its projection y1on the vertical axis varies sinusoidally, from a
maximum of ym1through zero to a minimum of ym1and then back to ym1. This
variation corresponds to the sinusoidal variation in the displacement y1of any
point along the string as the wave passes through that point. (All this is shown as
an animation with voiceover in WileyPLUS.)
When two waves travel along the same string in the same direction, we can
represent them and their resultant wave in a phasor diagram. The phasors in
Fig. 16-15erepresent the wave of Eq.16-55 and a second wave given by
y2(x,t)ym2sin(kx vtf). (16-56)
This second wave is phase-shifted from the first wave by phase constant f.
Because the phasors rotate at the same angular speed v, the angle between the
two phasors is always f. If fis a positive quantity, then the phasor for wave 2 lags
the phasor for wave 1 as they rotate, as drawn in Fig. 16-15e. If fis a negative
quantity, then the phasor for wave 2 leads the phasor for wave 1.
Because waves y1and y2have the same angular wave number kand angu-
lar frequency v, we know from Eqs. 16-51 and 16-52 that their resultant is of
the form
y(x,t)y
msin(kx vtb), (16-57)
16-6 PHASORS
After reading this module, you should be able to . . .
16.22 Using sketches, explain how a phasor can represent
the oscillations of a string element as a wave travels
through its location.
16.23 Sketch a phasor diagram for two overlapping waves
traveling together on a string, indicating their amplitudes
and phase difference on the sketch.
16.24 By using phasors, find the resultant wave of two trans-
verse waves traveling together along a string, calculating
the amplitude and phase and writing out the displacement
equation, and then displaying all three phasors in a phasor
diagram that shows the amplitudes, the leading or lagging,
and the relative phases.
●A wave y(x,t) can be represented with a phasor. This is a
vector that has a magnitude equal to the amplitude ymof the
wave and that rotates about an origin with an angular speed
equal to the angular frequency vof the wave. The projection
of the rotating phasor on a vertical axis gives the displace-
ment yof a point along the wave’s travel.
Learning Objectives
Key Idea
463
16-6 PHASORS
Figure 16-15 (a)–(d) A phasor of magnitude ym1rotating about an origin at angular speed vrepresents a sinusoidal wave.The phasor’s
projection y1on the vertical axis represents the displacement of a point through which the wave passes. (e) A second phasor, also of
angular speed vbut of magnitude ym2and rotating at a constant angle ffrom the first phasor,represents a second wave,with a phase
constant f.(f) The resultant wave is represented by the vector sum y
mof the two phasors.
φ
φ
y1
y
x
ym1
ω
ω
ω
y1ym1
y2ym2
y1ym1
y2
y'
ym2
y'
m
β
(a)
(e)(f)
y1= 0
y
x
ω
(b)
y1=ym1
y1
x
y
ω
(c)
ym1y1
x
y
ω
(d)
Zero projection,
zero displacement
Maximum negative projection The next crest is about to
move through the dot.
This is a snapshot of the
two phasors for two waves.
These are the
projections of
the two phasors.
Wave 1
This is the
projection of
the resultant
phasor.
Adding the two phasors as vectors
gives the resultant phasor of the
resultant wave.
Wave 2, delayed
by radians
φ
This projection matches this
displacement of the dot as
the wave moves through it.
A
464 CHAPTER 16 WAVES—I
where y
mis the amplitude of the resultant wave and bis its phase constant. To
find the values of y
mand b, we would have to sum the two combining waves, as
we did to obtain Eq. 16-51.To do this on a phasor diagram, we vectorially add the
two phasors at any instant during their rotation, as in Fig. 16-15fwhere phasor ym2
has been shifted to the head of phasor ym1. The magnitude of the vector sum
equals the amplitude y
min Eq. 16-57. The angle between the vector sum and the
phasor for y1equals the phase constant bin Eq. 16-57.
Note that, in contrast to the method of Module 16-5:
We can use phasors to combine waves even if their amplitudes are different.
we shall sum them by components. (They are called hori-
zontal and vertical components, because the symbols xand
yare already used for the waves themselves.) For the hori-
zontal components we have
y
mh ym1cos 0 ym2cos p/3
4.0 mm (3.0 mm) cos p/3 5.50 mm.
For the vertical components we have
y
mv ym1sin 0 ym2sin p/3
0(3.0 mm) sin p/3 2.60 mm.
Thus, the resultant wave has an amplitude of
6.1 mm (Answer)
and a phase constant of
(Answer)
From Fig. 16-16b, phase constant bis a positive angle rela-
tive to phasor 1.Thus, the resultant wave lags wave 1 in their
travel by phase constant b0.44 rad. From Eq. 16-57, we
can write the resultant wave as
y(x,t)(6.1 mm) sin(kx vt0.44 rad). (Answer)
btan12.60 mm
5.50 mm 0.44 rad.
y
m2(5.50 mm)2(2.60 mm)2
Sample Problem 16.05 Interference of two waves, same direction, phasors, any amplitudes
Two sinusoidal waves y1(x,t) and y2(x,t) have the same
wavelength and travel together in the same direction along
a string. Their amplitudes are ym14.0 mm and ym23.0
mm, and their phase constants are 0 and p/3 rad, respec-
tively. What are the amplitude y
mand phase constant bof
the resultant wave? Write the resultant wave in the form of
Eq. 16-57.
KEY IDEAS
(1) The two waves have a number of properties in com-
mon: Because they travel along the same string, they must
have the same speed v, as set by the tension and linear
density of the string according to Eq. 16-26. With the
same wavelength l, they have the same angular wave
number k(2p/l). Also, because they have the same
wave number kand speed v, they must have the same an-
gular frequency v(kv).
(2) The waves (call them waves 1 and 2) can be repre-
sented by phasors rotating at the same angular speed v
about an origin. Because the phase constant for wave 2 is
greater than that for wave 1 by p/3, phasor 2 must lag pha-
sor 1 by p/3 rad in their clockwise rotation, as shown in
Fig. 16-16a. The resultant wave due to the interference of
waves 1 and 2 can then be represented by a phasor that is
the vector sum of phasors 1 and 2.
Calculations: To simplify the vector summation, we drew
phasors 1 and 2 in Fig. 16-16aat the instant when phasor 1
lies along the horizontal axis. We then drew lagging
phasor 2 at positive angle p/3 rad. In Fig. 16-16b
we shifted phasor 2 so its tail is at the head of phasor 1.
Then we can draw the phasor y
mof the resultant
wave from the tail of phasor 1 to the head of phasor 2.
The phase constant bis the angle phasor y
mmakes with
phasor 1.
To find values for y
mand b, we can sum phasors 1 and
2 as vectors on a vector-capable calculator. However, here
Figure 16-16 (a) Two phasors of magnitudes ym1and ym2and with
phase difference p/3. (b) Vector addition of these phasors at any
instant during their rotation gives the magnitude y
mof the phasor
for the resultant wave.
π
/3
ym2
ym1
β π
ym2
/3
ym1
y'
m
y'
(
a
)(
b
)
Add the phasors
as vectors.
Additional examples, video, and practice available at WileyPLUS
465
16-7 STANDING WAVES AND RESONANCE
16-7 STANDING WAVES AND RESONANCE
After reading this module, you should be able to . . .
16.25 For two overlapping waves (same amplitude and
wavelength) that are traveling in opposite directions,
sketch snapshots of the resultant wave, indicating nodes
and antinodes.
16.26 For two overlapping waves (same amplitude and
wavelength) that are traveling in opposite directions, find
the displacement equation for the resultant wave and
calculate the amplitude in terms of the individual wave
amplitude.
16.27 Describe the SHM of a string element at an antinode
of a standing wave.
16.28 For a string element at an antinode of a standing wave,
write equations for the displacement, transverse velocity,
and transverse acceleration as functions of time.
16.29 Distinguish between “hard” and “soft” reflections of
string waves at a boundary.
16.30 Describe resonance on a string tied taut between two
supports, and sketch the first several standing wave
patterns, indicating nodes and antinodes.
16.31 In terms of string length, determine the wavelengths re-
quired for the first several harmonics on a string under tension.
16.32 For any given harmonic, apply the relationship between
frequency, wave speed, and string length.
●The interference of two identical sinusoidal waves moving
in opposite directions produces standing waves. For a string
with fixed ends, the standing wave is given by
y(x,t)[2ymsin kx] cos vt.
Standing waves are characterized by fixed locations of zero
displacement called nodes and fixed locations of maximum
displacement called antinodes.
●Standing waves on a string can be set up by reflection of
traveling waves from the ends of the string. If an end is fixed, it
must be the position of a node. This limits the frequencies at
which standing waves will occur on a given string. Each
possible frequency is a resonant frequency, and the
corresponding standing wave pattern is an oscillation mode.
For a stretched string of length Lwith fixed ends, the
resonant frequencies are
for n1,2,3,....
The oscillation mode corresponding to n1 is called the
fundamental mode or the first harmonic; the mode
corresponding to n2 is the second harmonic; and so on.
fv
lnv
2L,
Learning Objectives
Key Ideas
Standing Waves
In Module 16-5, we discussed two sinusoidal waves of the same wavelength and
amplitude traveling in the same direction along a stretched string. What if they
travel in opposite directions? We can again find the resultant wave by applying
the superposition principle.
Figure 16-17 suggests the situation graphically. It shows the two combin-
ing waves, one traveling to the left in Fig. 16-17a, the other to the right in
Fig. 16-17b. Figure 16-17cshows their sum, obtained by applying the superposition
Figure 16-17 (a) Five snapshots of a wave
traveling to the left,at the times t
indicated below part (c) (Tis the period
of oscillation).(b) Five snapshots of a
wave identical to that in (a) but traveling
to the right,at the same times t.(c)
Corresponding snapshots for the
superposition of the two waves on the
same string.At t0, T,and T,fully
constructive interference occurs because
of the alignment of peaks with peaks and
valleys with valleys.At tTand T,
fully destructive interference occurs
because of the alignment of peaks with
valleys.Some points (the nodes,marked
with dots) never oscillate;some points
(the antinodes) oscillate the most.
3
4
1
4
1
2
(a)
(b)
(c)
t= 0 t=T t =T
1
2t=T
3
4
t=T
1
4
xxxxx
As the waves move through each other,
some points never move and some move
the most.
466 CHAPTER 16 WAVES—I
principle graphically.The outstanding feature of the resultant wave is that there are
places along the string,called nodes, where the string never moves.Four such nodes
are marked by dots in Fig. 16-17c. Halfway between adjacent nodes are antinodes,
where the amplitude of the resultant wave is a maximum. Wave patterns such as
that of Fig. 16-17care called standing waves because the wave patterns do not
move left or right; the locations of the maxima and minima do not change.
If two sinusoidal waves of the same amplitude and wavelength travel in opposite
directions along a stretched string, their interference with each other produces a
standing wave.
Figure 16-18 The resultant wave of Eq.16-60
is a standing wave and is due to the
interference of two sinusoidal waves of the
same amplitude and wavelength that travel
in opposite directions.
Displacement
Magnitude
gives
amplitude
at position x
y'(x,t) = [2ym sin kx]cos t
ω
Oscillating
term
To analyze a standing wave, we represent the two waves with the equations
y1(x,t)ymsin(kx vt) (16-58)
and y2(x,t)ymsin(kx vt). (16-59)
The principle of superposition gives, for the combined wave,
y(x,t)y1(x,t)y2(x,t)ymsin(kx vt)ymsin(kx vt).
Applying the trigonometric relation of Eq. 16-50 leads to Fig. 16-18 and
y(x,t)[2ymsin kx] cos vt. (16-60)
This equation does not describe a traveling wave because it is not of the form of
Eq. 16-17. Instead, it describes a standing wave.
The quantity 2ymsin kx in the brackets of Eq. 16-60 can be viewed as the
amplitude of oscillation of the string element that is located at position x.
However, since an amplitude is always positive and sin kx can be negative, we
take the absolute value of the quantity 2ymsin kx to be the amplitude at x.
In a traveling sinusoidal wave, the amplitude of the wave is the same for all
string elements.That is not true for a standing wave, in which the amplitude varies
with position. In the standing wave of Eq. 16-60, for example, the amplitude is
zero for values of kx that give sin kx 0.Those values are
kx np, for n0,1,2,.... (16-61)
Substituting k2p/lin this equation and rearranging,we get
for n0,1,2,... (nodes), (16-62)
as the positions of zero amplitude—the nodes—for the standing wave of
Eq. 16-60. Note that adjacent nodes are separated by l/2, half a wavelength.
The amplitude of the standing wave of Eq. 16-60 has a maximum value of
2ym, which occurs for values of kx that give | sin kx |1.Those values are
,...
for n0,1,2,.... (16-63)
Substituting k2p/lin Eq. 16-63 and rearranging, we get
for n0,1,2,... (antinodes), (16-64)
as the positions of maximum amplitude—the antinodes—of the standing wave
of Eq. 16-60.Antinodes are separated by l/2 and are halfway between nodes.
Reflections at a Boundary
We can set up a standing wave in a stretched string by allowing a traveling wave
to be reflected from the far end of the string so that the wave travels back
x
n1
2
l
2,
(n1
2)p,
kx 1
2p,3
2p,5
2p
xnl
2,
467
16-7 STANDING WAVES AND RESONANCE
through itself. The incident (original) wave and the reflected wave can then be
described by Eqs. 16-58 and 16-59, respectively, and they can combine to form a
pattern of standing waves.
In Fig. 16-19, we use a single pulse to show how such reflections take place. In
Fig. 16-19a,the string is fixed at its left end.When the pulse arrives at that end,it ex-
erts an upward force on the support (the wall). By Newton’s third law, the support
exerts an opposite force of equal magnitude on the string.This second force gener-
ates a pulse at the support, which travels back along the string in the direction op-
posite that of the incident pulse. In a “hard” reflection of this kind, there must be a
node at the support because the string is fixed there. The reflected and incident
pulses must have opposite signs, so as to cancel each other at that point.
In Fig. 16-19b, the left end of the string is fastened to a light ring that is free to
slide without friction along a rod.When the incident pulse arrives, the ring moves
up the rod. As the ring moves, it pulls on the string, stretching the string and
producing a reflected pulse with the same sign and amplitude as the incident
pulse. Thus, in such a “soft” reflection, the incident and reflected pulses rein-
force each other, creating an antinode at the end of the string; the maximum
displacement of the ring is twice the amplitude of either of these two pulses.
Figuer 16-19 (a) A pulse incident from the
right is reflected at the left end of the
string,which is tied to a wall. Note that the
reflected pulse is inverted from the incident
pulse. (b) Here the left end of the string is
tied to a ring that can slide without friction
up and down the rod. Now the pulse is not
inverted by the reflection.
(a) (b)
There are two ways a
pulse can reflect from
the end of a string.
Checkpoint 5
Two waves with the same amplitude and wavelength interfere in three different
situations to produce resultant waves with the following equations:
(1) y(x,t)4 sin(5x4t)
(2) y(x,t)4 sin(5x) cos(4t)
(3) y(x,t)4 sin(5x4t)
In which situation are the two combining waves traveling (a) toward positive x,
(b) toward negative x, and (c) in opposite directions?
Standing Waves and Resonance
Consider a string, such as a guitar string, that is stretched between two clamps.
Suppose we send a continuous sinusoidal wave of a certain frequency along the
string, say, toward the right. When the wave reaches the right end, it reflects and
begins to travel back to the left.That left-going wave then overlaps the wave that
is still traveling to the right. When the left-going wave reaches the left end, it
reflects again and the newly reflected wave begins to travel to the right, over-
lapping the left-going and right-going waves. In short, we very soon have many
overlapping traveling waves, which interfere with one another.
For certain frequencies, the interference produces a standing wave pattern
(or oscillation mode) with nodes and large antinodes like those in Fig. 16-20.
Such a standing wave is said to be produced at resonance, and the string is said
to resonate at these certain frequencies, called resonant frequencies. If the string
Figure 16-20 Stroboscopic photographs reveal (imperfect) standing wave patterns on a
string being made to oscillate by an oscillator at the left end.The patterns occur at certain
frequencies of oscillation.
Richard Megna/Fundamental Photographs
Courtesy Thomas D. Rossing, Northern
Illinois University
Figure 16-22 One of many possible standing
wave patterns for a kettledrum head, made
visible by dark powder sprinkled on the
drumhead.As the head is set into oscilla-
tion at a single frequency by a mechanical
oscillator at the upper left of the photo-
graph, the powder collects at the nodes,
which are circles and straight lines in this
two-dimensional example.
Checkpoint 6
In the following series of resonant frequencies, one frequency (lower than 400 Hz)
is missing: 150, 225, 300, 375 Hz. (a) What is the missing frequency? (b) What is the
frequency of the seventh harmonic?
468 CHAPTER 16 WAVES—I
Figure 16-21 A string, stretched between two
clamps, is made to oscillate in standing
wave patterns. (a) The simplest possible
pattern consists of one loop, which refers to
the composite shape formed by the string
in its extreme displacements (the solid and
dashed lines). (b) The next simplest pattern
has two loops. (c) The next has three loops.
L
L =
λ
2
(a)
First harmonic
L =
λ
2
(b)2
λ
=
Second harmonic
L =
λ
2
(c)3
Third harmonic
is oscillated at some frequency other than a resonant frequency, a standing wave
is not set up. Then the interference of the right-going and left-going traveling
waves results in only small, temporary (perhaps even imperceptible) oscillations
of the string.
Let a string be stretched between two clamps separated by a fixed
distance L. To find expressions for the resonant frequencies of the string, we
note that a node must exist at each of its ends, because each end is fixed and
cannot oscillate. The simplest pattern that meets this key requirement is that
in Fig. 16-21a, which shows the string at both its extreme displacements (one
solid and one dashed, together forming a single “loop”). There is only one
antinode, which is at the center of the string. Note that half a wavelength
spans the length L, which we take to be the string’s length. Thus, for this
pattern, l/2 L. This condition tells us that if the left-going and right-going
traveling waves are to set up this pattern by their interference, they must have
the wavelength l2L.
A second simple pattern meeting the requirement of nodes at the fixed ends
is shown in Fig.16-21b.This pattern has three nodes and two antinodes and is said
to be a two-loop pattern. For the left-going and right-going waves to set it up,
they must have a wavelength lL.A third pattern is shown in Fig. 16-21c. It has
four nodes, three antinodes, and three loops, and the wavelength is lL.We could
continue this progression by drawing increasingly more complicated patterns. In
each step of the progression, the pattern would have one more node and one
more antinode than the preceding step, and an additional l/2 would be fitted into
the distance L.
Thus, a standing wave can be set up on a string of length Lby a wave with a
wavelength equal to one of the values
for n1,2,3,.... (16-65)
The resonant frequencies that correspond to these wavelengths follow from
Eq. 16-13:
for n1,2,3,.... (16-66)
Here vis the speed of traveling waves on the string.
Equation 16-66 tells us that the resonant frequencies are integer multiples of
the lowest resonant frequency, fv/2L, which corresponds to n1. The oscilla-
tion mode with that lowest frequency is called the fundamental mode or the first
harmonic. The second harmonic is the oscillation mode with n2, the third har-
monic is that with n3, and so on. The frequencies associated with these modes
are often labeled f1,f2,f3, and so on. The collection of all possible oscillation
modes is called the harmonic series, and nis called the harmonic number of the
nth harmonic.
For a given string under a given tension, each resonant frequency corre-
sponds to a particular oscillation pattern. Thus, if the frequency is in the audi-
ble range, you can hear the shape of the string. Resonance can also occur in
two dimensions (such as on the surface of the kettledrum in Fig. 16-22) and in
three dimensions (such as in the wind-induced swaying and twisting of a tall
building).
fv
nv
2L,
2L
n,
2
3
469
16-7 STANDING WAVES AND RESONANCE
(Answer)
Note that we get the same answer by substituting into Eq.
16-66:
(Answer)
Now note that this 806 Hz is not only the frequency of the
waves producing the fourth harmonic but also it is said to be
the fourth harmonic, as in the statement, “The fourth
harmonic of this oscillating string is 806 Hz.” It is also the
frequency of the string elements as they oscillate vertically
in the figure in simple harmonic motion, just as a block on a
vertical spring would oscillate in simple harmonic motion.
Finally, it is also the frequency of the sound you would hear
as the oscillating string periodically pushes against the air.
Transverse velocity: The displacement yof the string
element located at coordinate xis given by Eq. 16-67 as a
function of time t. The term cos vtcontains the dependence
on time and thus provides the “motion” of the standing
wave. The term 2ymsin kx sets the extent of the motion—
that is, the amplitude. The greatest amplitude occurs at an
antinode, where sin kx is 1 or 1 and thus the greatest
amplitude is 2ym. From Fig. 16-23, we see that 2ym4.00 mm,
which tells us that ym2.00 mm.
We want the transverse velocity—the velocity of a
string element parallel to the yaxis. To find it, we take the
time derivative of Eq. 16-67:
(16-69)
Here the term sin vtprovides the variation with time and
the term 2ymvsin kx provides the extent of that varia-
tion.We want the absolute magnitude of that extent:
To evaluate this for the element at x0.180 m, we first note
that ym2.00 mm, k2p/l2p/(0.400 m), and v
2pf2p(806.2 Hz). Then the maximum speed of the
element at x0.180 m is
um2ymv sin kx .
[2ym
sin kx] sin
t.
u(x,t)y
t
t [(2ym sin kx) cos vt]
806 Hz.
fnv
2L4322.49 m/s
2(0.800 m)
806.2 Hz 806 Hz.
Figure 16-23 shows resonant oscillation of a string of mass
m2.500 g and length L0.800 m and that is under tension
t325.0 N. What is the wavelength lof the transverse
waves producing the standing wave pattern, and what is the
harmonic number n? What is the frequency fof the trans-
verse waves and of the oscillations of the moving string ele-
ments? What is the maximum magnitude of the transverse
velocity umof the element oscillating at coordinate x0.180 m?
At what point during the element’s oscillation is the trans-
verse velocity maximum?
KEY IDEAS
(1) The traverse waves that produce a standing wave pattern
must have a wavelength such that an integer number nof
half-wavelengths fit into the length Lof the string. (2) The
frequency of those waves and of the oscillations of the string
elements is given by Eq. 16-66 (fnv/2L). (3) The displace-
ment of a string element as a function of position xand time
tis given by Eq. 16-60:
y(x,t)[2ymsin kx] cos vt. (16-67)
Wavelength and harmonic number: In Fig. 16-23, the solid
line, which is effectively a snapshot (or freeze-frame) of the
oscillations, reveals that 2 full wavelengths fit into the length
L0.800 m of the string.Thus,we have
or (16-68)
(Answer)
By counting the number of loops (or half-wavelengths) in
Fig.16-23, we see that the harmonic number is
n4. (Answer)
We also find n4 by comparing Eqs. 16-68 and 16-65 (l
2L/n).Thus, the string is oscillating in its fourth harmonic.
Frequency: We can get the frequency fof the transverse waves
from Eq. 16-13 (vlf) if we first find the speed vof the waves.
That speed is given by Eq.16-26,but we must substitute m/Lfor
the unknown linear density m.We obtain
After rearranging Eq. 16-13, we write
fv
l322.49 m/s
0.400 m
A(325 N)(0.800 m)
2.50 103 kg 322.49 m/s.
vAt
mAt
m/LAtL
m
0.800 m
20.400 m.
L
2.
2 lL,
Figure 16-23 Resonant oscillation of a string under tension.
0.800
x (m)
y
8.00 mm
0
Sample Problem 16.06 Resonance of transverse waves, standing waves, harmonics
Transverse and Longitudinal Waves Mechanical waves
can exist only in material media and are governed by Newton’s
laws. Transverse mechanical waves, like those on a stretched string,
are waves in which the particles of the medium oscillate perpendi-
cular to the wave’s direction of travel.Waves in which the particles
of the medium oscillate parallel to the wave’s direction of travel
are longitudinal waves.
Sinusoidal Waves A sinusoidal wave moving in the positive
direction of an xaxis has the mathematical form
y(x,t)ymsin(kx vt), (16-2)
where ymis the amplitude of the wave, kis the angular wave number,
vis the angular frequency, and kx vtis the phase. The wavelength
lis related to kby
(16-5)
The period Tand frequency fof the wave are related to vby
(16-9)
Finally,the wave speed vis related to these other parameters by
(16-13)
Equation of a Traveling Wave Any function of the form
y(x,t)h(kx vt) (16-17)
can represent a traveling wave with a wave speed given by Eq. 16-13
and a wave shape given by the mathematical form of h.The plus sign
denotes a wave traveling in the negative direction of the xaxis, and
the minus sign a wave traveling in the positive direction.
Wave Speed on Stretched String The speed of a wave on
a stretched string is set by properties of the string. The speed on a
string with tension tand linear density mis
(16-26)
Power The average power of, or average rate at which energy is
transmitted by, a sinusoidal wave on a stretched string is given by
Pavg (16-33)
1
2mvv2y2
m.
vAt
m.
v
k
T
f.
2
f1
T.
k2
.
Review & Summary
Superposition of Waves When two or more waves traverse
the same medium, the displacement of any particle of the medium is
the sum of the displacements that the individual waves would give it.
Interference of Waves Two sinusoidal waves on the same
string exhibit interference, adding or canceling according to the prin-
ciple of superposition. If the two are traveling in the same direction
and have the same amplitude ymand frequency (hence the same
wavelength) but differ in phase by a phase constant f, the result is a
single wave with this same frequency:
y(x,t)[2ymcos f] sin(kx vtf). (16-51)
If f0, the waves are exactly in phase and their interference is
fully constructive; if fprad, they are exactly out of phase and
their interference is fully destructive.
Phasors A wave y(x,t) can be represented with a phasor. This
is a vector that has a magnitude equal to the amplitude ymof the
wave and that rotates about an origin with an angular speed equal
to the angular frequency vof the wave.The projection of the rotat-
ing phasor on a vertical axis gives the displacement yof a point
along the wave’s travel.
Standing Waves The interference of two identical sinusoidal
waves moving in opposite directions produces standing waves. For
a string with fixed ends, the standing wave is given by
y(x,t)[2ymsin kx] cos vt. (16-60)
Standing waves are characterized by fixed locations of zero dis-
placement called nodes and fixed locations of maximum displace-
ment called antinodes.
Resonance Standing waves on a string can be set up by
reflection of traveling waves from the ends of the string. If an end
is fixed, it must be the position of a node. This limits the frequen-
cies at which standing waves will occur on a given string. Each pos-
sible frequency is a resonant frequency, and the corresponding
standing wave pattern is an oscillation mode. For a stretched string
of length Lwith fixed ends, the resonant frequencies are
for n1,2,3,.... (16-66)
The oscillation mode corresponding to n1 is called the funda-
mental mode or the first harmonic; the mode corresponding to
n2 is the second harmonic; and so on.
fv
nv
2L,
1
2
1
2
470 CHAPTER 16 WAVES—I
(Answer)6.26 m/s.
sin
2p
0.400 m (0.180 m)
um
2(2.00 103 m)(2p)(806.2 Hz)
Additional examples, video, and practice available at WileyPLUS
To determine when the string element has this maxi-
mum speed, we could investigate Eq. 16-69.However,a little
thought can save a lot of work. The element is undergoing
SHM and must come to a momentary stop at its extreme
upward position and extreme downward position. It has the
greatest speed as it zips through the midpoint of its oscilla-
tion, just as a block does in a block–spring oscillator.
10 If you set up the seventh harmonic on a string, (a) how many
nodes are present, and (b) is there a node, antinode, or some inter-
mediate state at the midpoint? If you next set up the sixth harmonic,
(c) is its resonant wavelength longer or shorter than that for the sev-
enth harmonic, and (d) is
the resonant frequency
higher or lower?
11 Figure 16-28 shows
phasor diagrams for three
situations in which two
waves travel along the
same string. All six waves
have the same amplitude. Rank the situations according to the am-
plitude of the net wave on the string,greatest first.
471
QUESTIONS
Questions
1The following four waves are sent along strings with the same
linear densities (xis in meters and tis in seconds). Rank the waves
according to (a) their wave speed and (b) the tension in the strings
along which they travel, greatest first:
(1) y1(3 mm) sin(x3t),
(2) y2(6 mm) sin(2xt),
2In Fig. 16-24, wave 1 consists of a rectangular peak of height 4 units
and width d, and a rectangular valley of depth 2 units and width d.The
wave travels rightward along an xaxis. Choices 2, 3, and 4 are similar
waves,with the same heights,depths, and widths,that will travel leftward
along that axis and through wave 1. Right-going wave 1 and one of the
left-going waves will interfere as they pass through each other. With
which left-going wave will the interference give, for an instant, (a) the
deepest valley,(b) a flat line,and (c) a flat peak 2dwide?
6The amplitudes and phase differences for four pairs of
waves of equal wavelengths are (a) 2 mm, 6 mm, and prad;
(b) 3 mm, 5 mm, and prad; (c) 7 mm, 9 mm, and prad; (d) 2 mm,
2 mm, and 0 rad. Each pair travels in the same direction along the
same string. Without written calculation, rank the four pairs ac-
cording to the amplitude of their resultant wave, greatest first.
(Hint: Construct phasor diagrams.)
7A sinusoidal wave is sent along a cord under tension, transport-
ing energy at the average rate of Pavg,1.Two waves, identical to that
first one, are then to be sent along the cord with a phase difference
fof either 0, 0.2 wavelength, or 0.5 wavelength. (a) With only men-
tal calculation, rank those choices of faccording to the average
rate at which the waves will transport energy, greatest first. (b) For
the first choice of f, what is the average rate in terms of Pavg,1?
8(a) If a standing wave on a string is given by
y(t)(3 mm) sin(5x) cos(4t),
is there a node or an antinode of the oscillations of the string at
x0? (b) If the standing wave is given by
y(t)(3 mm) sin(5xp/2) cos(4t),
is there a node or an antinode at x0?
9Strings Aand Bhave identical lengths and linear densities, but
string Bis under greater tension than string A. Figure 16-27 shows
four situations, (a) through (d), in which standing wave patterns
exist on the two strings. In which situations is there the possibility
that strings Aand Bare oscillating at the same resonant frequency?
(3) y3(1 mm) sin(4xt),
(4) y4(2 mm) sin(x2t).
(3) (4)
(1) (2)
Figure 16-24 Question 2.
3Figure 16-25agives a snapshot of a wave traveling in the direc-
tion of positive xalong a string under tension. Four string elements
are indicated by the lettered points. For each of those elements, de-
termine whether, at the instant of the snapshot, the element is
moving upward or downward or is momentarily at rest. (Hint:
Imagine the wave as it moves through the four string elements, as if
you were watching a video of the wave as it traveled rightward.)
Figure 16-25bgives the displacement of a string element located
at, say, x0 as a function of time. At the lettered times, is the
element moving upward or downward or is it momentarily at rest?
Figure 16-25 Question 3.
y y
x t
he
f
a d
b c
g
(a) (b)
4Figure 16-26 shows three waves that
are separately sent along a string that is
stretched under a certain tension along
an xaxis. Rank the waves according to
their (a) wavelengths, (b) speeds, and
(c) angular frequencies,greatest first.
5If you start with two sinusoidal
waves of the same amplitude travel-
ing in phase on a string and then
somehow phase-shift one of them by
5.4 wavelengths, what type of interference will occur on the string?
Figure 16-26 Question 4.
y
x
3
2
1
Figure 16-27 Question 9.
(a)
(b)
(c)
(
d
)
S
tr
i
ng A
S
tr
i
ng B
Figure 16-28 Question 11.
(
a
) (
b
)(
c
)
472 CHAPTER 16 WAVES—I
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Module 16-1 Transverse Waves
•1 If a wave y(x,t)(6.0 mm) sin(kx (600 rad/s)tf) travels
along a string, how much time does any given point on the string take
to move between displacements y2.0 mm and y2.0 mm?
•2 A human wave. During
sporting events within large, densely
packed stadiums, spectators will
send a wave (or pulse) around the
stadium (Fig. 16-29). As the wave
reaches a group of spectators, they
stand with a cheer and then sit. At
any instant, the width wof the wave
is the distance from the leading edge (people are just about to stand)
to the trailing edge (people have just sat down). Suppose a human
wave travels a distance of 853 seats around a stadium in 39 s, with
spectators requiring about 1.8 s to respond to the wave’s passage by
standing and then sitting.What are (a) the wave speed v(in seats per
second) and (b) width w(in number of seats)?
•3 A wave has an angular frequency of 110 rad/s and a wave-
length of 1.80 m. Calculate (a) the angular wave number and
(b) the speed of the wave.
•4 A sand scorpion can de-
tect the motion of a nearby beetle
(its prey) by the waves the motion
sends along the sand surface (Fig.
16-30). The waves are of two types:
transverse waves traveling at
and longitudinal waves
traveling at . If a sud-
den motion sends out such waves, a
scorpion can tell the distance of the
beetle from the difference tin the
arrival times of the waves at its leg
nearest the beetle. If t4.0 ms,
what is the beetle’s distance?
•5 A sinusoidal wave travels along
a string. The time for a particular
point to move from maximum displacement to zero is 0.170 s.What
are the (a) period and (b) frequency? (c) The wavelength is 1.40 m;
what is the wave speed?
••6 A sinusoidal wave travels
along a string under tension.
Figure 16-31 gives the slopes
along the string at time t0.The
scale of the xaxis is set by xs
0.80 m. What is the amplitude of
the wave?
••7 A transverse sinusoidal wave is moving along a string in the
positive direction of an xaxis with a speed of 80 m/s. At t0, the
string particle at x0 has a transverse displacement of 4.0 cm
from its equilibrium position and is not moving. The maximum
vl150 m/s
vt50 m/s
transverse speed of the string particle at x0 is 16 m/s. (a) What is
the frequency of the wave? (b) What is the wavelength of the
wave? If y(x,t)ymsin(kx vtf) is the form of the wave
equation, what are (c) ym, (d) k, (e) v,(f)f, and (g) the correct
choice of sign in front of v?
••8 Figure 16-32 shows the trans-
verse velocity uversus time tof the
point on a string at x0, as a wave
passes through it.The scale on the ver-
tical axis is set by us4.0 m/s. The
wave has the generic form y(x,t)
ymsin(kx vtf). What then is f?
(Caution: A calculator does not always
give the proper inverse trig function, so
check your answer by substituting it
and an assumed value of vinto y(x,t) and then plotting the function.)
••9 A sinusoidal wave mov-
ing along a string is shown
twice in Fig. 16-33, as crest A
travels in the positive direc-
tion of an xaxis by distance
d6.0 cm in 4.0 ms. The
tick marks along the axis are
separated by 10 cm; height
H6.00 mm. The equation
for the wave is in the form
y(x,t)ymsin(kx vt), so
what are (a) ym, (b) k, (c) v, and (d) the correct choice of sign in
front of v?
••10 The equation of a transverse wave traveling along a very
long string is y6.0 sin(0.020px4.0pt), where xand yare ex-
pressed in centimeters and tis in seconds. Determine (a) the ampli-
tude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the
direction of propagation of the wave, and (f) the maximum trans-
verse speed of a particle in the string. (g) What is the transverse
displacement at x3.5 cm when t
0.26 s?
••11 A sinusoidal transverse wave
of wavelength 20 cm travels along a
string in the positive direction of an
xaxis. The displacement yof the
string particle at x0 is given in
Fig. 16-34 as a function of time t. The scale of the vertical axis is
set by ys4.0 cm. The wave equation is to be in the form
y(x,t)ymsin(kx vtf). (a) At t0, is a plot of yversus xin
the shape of a positive sine function or a negative sine function?
What are (b) ym, (c) k, (d) v, (e) f, (f) the sign in front of v, and (g)
the speed of the wave? (h) What is the transverse velocity of the par-
ticle at x0 when t5.0 s?
••12 The function y(x,t)(15.0 cm) cos(px15pt), with xin
meters and tin seconds, describes a wave on a taut string. What is
w
v
Figure 16-29 Problem 2.
Figure 16-30 Problem 4.
d
v
l
v
l
v
t
v
t
Beetle
Figure 16-31 Problem 6.
00
–0.2
0.2
Slope
x (m)
xs
us
–us
u (m/s)
t
Figure 16-32 Problem 8.
Hx
A
y
d
Figure 16-33 Problem 9.
y
(
cm
)
ys
–ys
t (s)
100
Figure 16-34 Problem 11.
end at x0 and the other at x10.0 m. At time t0, pulse 1 is
sent along the wire from the end at x10.0 m. At time t30.0
ms, pulse 2 is sent along the wire from the end at x0.At what po-
sition xdo the pulses begin to meet?
••22 A sinusoidal wave is traveling on a string with speed 40 cm/s.
The displacement of the particles of the string at x10 cm varies
with time according to y(5.0 cm) sin[1.0 (4.0 s1)t].The linear
density of the string is 4.0 g/cm.What
are (a) the frequency and (b) the
wavelength of the wave? If the wave
equation is of the form y(x,t)
ymsin(kx vt), what are (c) ym, (d) k,
(e) v, and (f) the correct choice of
sign in front of v? (g) What is the ten-
sion in the string?
••23 A sinusoidal trans-
verse wave is traveling along a string in
the negative direction of an xaxis.
Figure 16-35 shows a plot of the dis-
ILWSSM
473
PROBLEMS
the transverse speed for a point on the string at an instant when
that point has the displacement y12.0 cm?
••13 A sinusoidal wave of frequency 500 Hz has a speed of
350 m/s. (a) How far apart are two points that differ in phase by p/3
rad? (b) What is the phase difference between two displacements
at a certain point at times 1.00 ms apart?
Module 16-2 Wave Speed on a Stretched String
•14 The equation of a transverse wave on a string is
y(2.0 mm) sin[(20 m1)x(600 s1)t].
The tension in the string is 15 N. (a) What is the wave speed? (b)
Find the linear density of this string in grams per meter.
•15 A stretched string has a mass per unit length of
5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string
has an amplitude of 0.12 mm and a frequency of 100 Hz and is
traveling in the negative direction of an xaxis. If the wave equation
is of the form y(x,t)ymsin(kx vt), what are (a) ym, (b) k, (c) v,
and (d) the correct choice of sign in front of v?
•16 The speed of a transverse wave on a string is 170 m/s when
the string tension is 120 N. To what value must the tension be
changed to raise the wave speed to 180 m/s?
•17 The linear density of a string is 1.6 104kg/m.A transverse
wave on the string is described by the equation
y(0.021 m) sin[(2.0 m1)x(30 s1)t].
What are (a) the wave speed and (b) the tension in the string?
•18 The heaviest and lightest strings on a certain violin have lin-
ear densities of 3.0 and 0.29 g/m. What is the ratio of the diameter
of the heaviest string to that of the lightest string,assuming that the
strings are of the same material?
•19 What is the speed of a transverse wave in a rope of
length 2.00 m and mass 60.0 g under a tension of 500 N?
•20 The tension in a wire clamped at both ends is doubled with-
out appreciably changing the wire’s length between the clamps.
What is the ratio of the new to the old wave speed for transverse
waves traveling along this wire?
••21 A 100 g wire is held under a tension of 250 N with one
ILW
SSM
WWWSSM
ILW
placement as a function of position at time t0; the scale of the
yaxis is set by ys4.0 cm. The string tension is 3.6 N, and its lin-
ear density is 25 g/m. Find the (a) amplitude, (b) wavelength,
(c) wave speed, and (d) period of the wave. (e) Find the maxi-
mum transverse speed of a particle in the string. If the wave is of
the form y(x,t)ymsin(kx vtf), what are (f) k, (g) v, (h)
f, and (i) the correct choice of sign
in front of v?
•••24 In Fig. 16-36a, string 1 has a
linear density of 3.00 g/m, and
string 2 has a linear density of 5.00
g/m. They are under tension due to
the hanging block of mass M500
g. Calculate the wave speed on (a)
string 1 and (b) string 2. (Hint:
When a string loops halfway
around a pulley, it pulls on the pul-
ley with a net force that is twice the
tension in the string.) Next the
block is divided into two blocks
(with M1M2M) and the appa-
ratus is rearranged as shown in
Fig. 16-36b. Find (c) M1and (d) M2
such that the wave speeds in the
two strings are equal.
•••25 A uniform rope of mass m
and length Lhangs from a ceiling.
(a) Show that the speed of a trans-
verse wave on the rope is a function
of y, the distance from the lower end, and is given by v(b)
Show that the time a transverse wave takes to travel the length of
the rope is given by
Module 16-3 Energy and Power of a Wave Traveling
Along a String
•26 A string along which waves can travel is 2.70 m long and has
a mass of 260 g. The tension in the string is 36.0 N. What must be
the frequency of traveling waves of amplitude 7.70 mm for the av-
erage power to be 85.0 W?
••27 A sinusoidal wave is sent along a string with a linear
density of 2.0 g/m. As it travels, the kinetic energies of
the mass elements along the string vary. Figure 16-37agives the
rate dK/dt at which kinetic energy passes through the string ele-
ments at a particular instant, plotted as a function of distance x
along the string. Figure 16-37bis similar except that it gives the
rate at which kinetic energy passes through a particular mass ele-
ment (at a particular location), plotted as a function of time t.For
both figures, the scale on the vertical (rate) axis is set by Rs10 W.
What is the amplitude of the wave?
t21L/g.
1gy.
M1M2
M
String 1
String 2
Knot
(a)
(b)
String 1
String 2
Figure 16-36 Problem 24.
dK/dt (W)
Rs
0 0.1
x (m)
0.2
dK/dt (W)
Rs
0 1
t (ms)
(a) (b)
2
Figure 16-37 Problem 27.
Figure 16-35 Problem 23.
y (cm)
x (cm)
ys
0
–ys
20 40
474 CHAPTER 16 WAVES—I
Module 16-4 The Wave Equation
•28 Use the wave equation to find the speed of a wave given by
y(x,t)(3.00 mm) sin[(4.00 m1)x(7.00 s1)t].
••29 Use the wave equation to find the speed of a wave given by
y(x,t)(2.00 mm)[(20 m1)x(4.0 s1)t]0.5.
•••30 Use the wave equation to find the speed of a wave given in
terms of the general function h(x,t):
y(x,t)(4.00 mm) h[(30 m1)x(6.0 s1)t].
Module 16-5 Interference of Waves
•31 Two identical traveling waves, moving in the same di-
rection, are out of phase by p/2 rad. What is the amplitude of the
resultant wave in terms of the common amplitude ymof the two
combining waves?
•32 What phase difference between two identical traveling
waves, moving in the same direction along a stretched string, re-
sults in the combined wave having an amplitude 1.50 times that of
the common amplitude of the two combining waves? Express
your answer in (a) degrees, (b) radians,and (c) wavelengths.
••33 Two sinusoidal waves with
the same amplitude of 9.00 mm and
the same wavelength travel together
along a string that is stretched along
an xaxis. Their resultant wave is
shown twice in Fig. 16-38, as valley A
travels in the negative direction of
the xaxis by distance d56.0 cm in
8.0 ms.The tick marks along the axis
are separated by 10 cm, and height
His 8.0 mm. Let the equation for
one wave be of the form y(x,t)ymsin(kx vtf1), where
f10 and you must choose the correct sign in front of v. For the
equation for the other wave, what are (a) ym, (b) k, (c) v, (d) f2,
and (e) the sign in front of v?
•••34 A sinusoidal wave of angular frequency 1200 rad/s
and amplitude 3.00 mm is sent along a cord with linear density
2.00 g/m and tension 1200 N. (a) What is the average rate at
SSM
••37 These two waves travel along the same string:
y1(x,t)(4.60 mm) sin(2px400pt)
y2(x,t)(5.60 mm) sin(2px400pt0.80prad).
What are (a) the amplitude and (b) the phase angle (relative to
wave 1) of the resultant wave? (c) If a third wave of amplitude
5.00 mm is also to be sent along the string in the same direction as
the first two waves, what should be its phase angle in order to
maximize the amplitude of the new resultant wave?
••38 Two sinusoidal waves of the same frequency are to be sent
in the same direction along a taut string. One wave has an ampli-
tude of 5.0 mm, the other 8.0 mm. (a) What phase difference f1be-
tween the two waves results in the smallest amplitude of the result-
ant wave? (b) What is that smallest amplitude? (c) What phase
difference f2results in the largest amplitude of the resultant
wave? (d) What is that largest amplitude? (e) What is the resultant
amplitude if the phase angle is (f1f2)/2?
••39 Two sinusoidal waves of the same period, with amplitudes of
5.0 and 7.0 mm, travel in the same direction along a stretched
string; they produce a resultant wave with an amplitude of 9.0 mm.
The phase constant of the 5.0 mm wave is 0.What is the phase con-
stant of the 7.0 mm wave?
Module 16-7 Standing Waves and Resonance
•40 Two sinusoidal waves with identical wavelengths and
amplitudes travel in opposite directions along a string with a speed
of 10 cm/s. If the time interval between instants when the string is
flat is 0.50 s, what is the wavelength of the waves?
•41 A string fixed at both ends is 8.40 m long and has a
mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscil-
lating. (a) What is the speed of the waves on the string? (b) What is
the longest possible wavelength for a standing wave? (c) Give the
frequency of that wave.
•42 A string under tension tioscillates in the third harmonic at fre-
quency f3, and the waves on the string have wavelength l3. If the ten-
sion is increased to tf4tiand the string is again made to oscillate in
the third harmonic, what then are (a) the frequency of oscillation in
terms of f3and (b) the wavelength of the waves in terms of l3?
•43 What are (a) the lowest frequency, (b) the sec-
ond lowest frequency, and (c) the third lowest frequency for stand-
ing waves on a wire that is 10.0 m long, has a mass of 100 g, and is
stretched under a tension of 250 N?
•44 A 125 cm length of string has mass 2.00 g and tension 7.00 N.
(a) What is the wave speed for this string? (b) What is the lowest
resonant frequency of this string?
•45 A string that is stretched between fixed supports
separated by 75.0 cm has resonant frequencies of 420 and 315 Hz,
with no intermediate resonant frequencies.What are (a) the lowest
resonant frequency and (b) the wave speed?
•46 String Ais stretched between two clamps separated by dis-
tance L. String B, with the same linear density and under the same
tension as string A, is stretched between two clamps separated by
distance 4L. Consider the first eight harmonics of string B.For
which of these eight harmonics of B(if any) does the frequency
match the frequency of (a) A’s first harmonic, (b) A’s second har-
monic,and (c) A’s third harmonic?
•47 One of the harmonic frequencies for a particular string under
tension is 325 Hz. The next higher harmonic frequency is 390 Hz.
ILWSSM
WWWSSM
SSM
Hx
A
d
y
Figure 16-38 Problem 33.
the same direction along a string. If ym13.0 cm, ym24.0 cm,
f10,and f2p/2 rad, what is the amplitude of the resultant wave?
••36 Four waves are to be sent along the same string, in the same
direction:
y1(x,t)(4.00 mm) sin(2px400pt)
y2(x,t)(4.00 mm) sin(2px400pt0.7p)
y3(x,t)(4.00 mm) sin(2px400ptp)
y4(x,t)(4.00 mm) sin(2px400pt1.7p).
What is the amplitude of the resultant wave?
which energy is transported by the wave to the opposite end of the
cord? (b) If, simultaneously, an identical wave travels along an adja-
cent, identical cord, what is the total average rate at which energy is
transported to the opposite ends of the two cords by the waves? If, in-
stead, those two waves are sent along the same cord simultaneously,
what is the total average rate at which they transport energy when
their phase difference is (c) 0,(d) 0.4prad, and (e) prad?
Module 16-6 Phasors
•35 Two sinusoidal waves of the same frequency travel in
SSM
475
PROBLEMS
What harmonic frequency is next higher after the harmonic fre-
quency 195 Hz?
•48 If a transmission line in a cold climate collects ice, the
increased diameter tends to cause vortex formation in a passing
wind. The air pressure variations in the vortexes tend to cause the
line to oscillate (gallop), especially if the frequency of the varia-
tions matches a resonant frequency of the line. In long lines, the
resonant frequencies are so close that almost any wind speed can
set up a resonant mode vigorous enough to pull down support tow-
ers or cause the line to short out with an adjacent line. If a transmis-
sion line has a length of 347 m, a linear density of 3.35 kg/m, and a
tension of 65.2 MN, what are (a) the frequency of the fundamental
mode and (b) the frequency difference between successive modes?
•49 A nylon guitar string has a
linear density of 7.20 g/m and is under a
tension of 150 N.The fixed supports are
distance D90.0 cm apart. The string
is oscillating in the standing wave pat-
tern shown in Fig. 16-39. Calculate the (a) speed, (b) wavelength, and
(c) frequency of the traveling waves whose superposition gives this
standing wave.
••50 For a particular transverse
standing wave on a long string, one
of the antinodes is at x0 and an
adjacent node is at x0.10 m. The
displacement y(t) of the string parti-
cle at x0 is shown in Fig. 16-40,
where the scale of the yaxis is set by
ys4.0 cm. When t0.50 s, what is
the displacement of the string particle
at (a) x0.20 m and (b) x0.30 m?
What is the transverse velocity of the string particle at x0.20 m at
(c) t0.50 s and (d) t1.0 s? (e) Sketch the standing wave at t
0.50 s for the range x0 to x0.40 m.
••51 Two waves are generated on a string of length
3.0 m to produce a three-loop standing wave with an amplitude of
1.0 cm. The wave speed is 100 m/s. Let the equation for one of the
waves be of the form y(x,t)ymsin(kx vt). In the equation for the
other wave, what are (a) ym, (b) k, (c) v, and (d) the sign in front of v?
••52 A rope, under a tension of 200 N and fixed at both ends, os-
cillates in a second-harmonic standing wave pattern. The displace-
ment of the rope is given by
y(0.10 m)(sin px/2) sin 12pt,
where x0 at one end of the rope, xis in meters, and tis in sec-
onds. What are (a) the length of the rope, (b) the speed of the
waves on the rope, and (c) the mass of the rope? (d) If the rope os-
cillates in a third-harmonic standing wave pattern, what will be the
period of oscillation?
••53 A string oscillates according to the equation
What are the (a) amplitude and (b) speed of the two waves
(identical except for direction of travel) whose superposition
gives this oscillation? (c) What is the distance between nodes?
(d) What is the transverse speed of a particle of the string at the
position x1.5 cm when ts?
9
8
y(0.50 cm) sin
p
3 cm1
x cos[(40p
s1)t].
WWWSSM
ILW
••54 Two sinusoidal waves
with the same amplitude and
wavelength travel through each
other along a string that is
stretched along an xaxis. Their
resultant wave is shown twice
in Fig. 16-41, as the antinode A
travels from an extreme up-
ward displacement to an ex-
treme downward displacement
in 6.0 ms. The tick marks along
the axis are separated by 10 cm; height His 1.80 cm. Let the equation
for one of the two waves be of the form y(x,t)ymsin(kx vt). In
the equation for the other wave, what are (a) ym, (b) k, (c) v, and (d)
the sign in front of v?
••55 The following two waves are sent in opposite directions
on a horizontal string so as to create a standing wave in a vertical
plane:
y1(x,t)(6.00 mm) sin(4.00px400pt)
y2(x,t)(6.00 mm) sin(4.00px400pt),
with xin meters and tin seconds.An antinode is located at point A.
In the time interval that point takes to move from maximum up-
ward displacement to maximum downward displacement, how far
does each wave move along the string?
••56 A standing wave pattern on a string is described by
y(x,t)0.040 (sin 5px)(cos 40pt),
where xand yare in meters and tis in seconds. For x0, what is
the location of the node with the (a) smallest, (b) second small-
est, and (c) third smallest value of x? (d) What is the period of the
oscillatory motion of any (nonnode) point? What are the (e)
speed and (f) amplitude of the two traveling waves that interfere
to produce this wave? For t0, what are the (g) first, (h) second,
and (i) third time that all points on the string have zero trans-
verse velocity?
••57 A generator at one end of a very long string creates a wave
given by
and a generator at the other end creates the wave
Calculate the (a) frequency, (b) wavelength, and (c) speed of each
wave. For x0, what is the location of the node having the (d)
smallest, (e) second smallest, and (f) third smallest value of x? For
x0, what is the location of the antinode having the (g) smallest,
(h) second smallest, and (i) third smallest value of x?
••58 In Fig. 16-42, a string, tied to a sinusoidal oscillator at P
and running over a support at Q, is stretched by a block of mass m.
Separation L1.20 m, linear density m 1.6 g/m, and the oscillator
y(6.0 cm) cos p
2 [(2.00 m1)x(8.00 s1)t].
y(6.0 cm) cos p
2 [(2.00 m1)x(8.00 s1)t],
D
Figure 16-39 Problem 49.
0.5 1.5 2.0
–
y
s
ys
0
y (cm)
t(s)
Figure 16-40 Problem 50.
Figure 16-41 Problem 54.
Hx
A
A
y
PQ
m
Oscillator
L
Figure 16-42 Problems 58 and 60.
476 CHAPTER 16 WAVES—I
sign), and (d) wavelength of the wave. (e) Find the maximum
transverse speed of a particle in the string.
66 Figure 16-44 shows the dis-
placement yversus time tof the
point on a string at x0, as a
wave passes through that point.
The scale of the yaxis is set by
ys6.0 mm. The wave is given
by y(x,t)ymsin(kx vtf).
What is f? (Caution: A calculator
does not always give the proper
inverse trig function, so check your answer by substituting it and an
assumed value of vinto y(x,t) and then plotting the function.)
67 Two sinusoidal waves, identical except for phase, travel in
the same direction along a string, producing the net wave
y(x,t)(3.0 mm) sin(20x4.0t0.820 rad), with xin meters
and tin seconds. What are (a) the wavelength lof the two waves,
(b) the phase difference between them, and (c) their amplitude ym?
68 A single pulse, given by h(x5.0t), is shown in Fig. 16-45
for t0. The scale of the vertical
axis is set by hs2. Here xis in
centimeters and tis in seconds.
What are the (a) speed and (b) di-
rection of travel of the pulse? (c)
Plot h(x5t) as a function of xfor
t2 s. (d) Plot h(x5t) as a func-
tion of tfor x10 cm.
69 Three sinusoidal waves of the same frequency travel
along a string in the positive direction of an xaxis. Their
amplitudes are y1,y1/2, and y1/3, and their phase constants
are 0, p/2, and p, respectively. What are the (a) amplitude and
(b) phase constant of the resultant wave? (c) Plot the wave
form of the resultant wave at t0, and discuss its behavior as t
increases.
70 Figure 16-46 shows
transverse acceleration ayversus
time tof the point on a string at
x0, as a wave in the form of
y(x, t)ymsin(kx vtf)
passes through that point. The
scale of the vertical axis is set
by as400 m/s2. What is f?
(Caution: A calculator does not
always give the proper inverse trig function, so check your answer by
substituting it and an assumed value of vinto y(x,t) and then plotting
the function.)
71 A transverse sinusoidal wave is generated at one end of a
long, horizontal string by a bar that moves up and down through
a distance of 1.00 cm. The motion is continuous and is repeated
regularly 120 times per second. The string has linear density 120
g/m and is kept under a tension of 90.0 N. Find the maximum
value of (a) the transverse speed uand (b) the transverse com-
ponent of the tension t.
(c) Show that the two maximum values calculated above
occur at the same phase values for the wave. What is the trans-
verse displacement yof the string at these phases? (d) What is the
maximum rate of energy transfer along the string? (e) What is the
transverse displacement ywhen this maximum transfer occurs?
(f) What is the minimum rate of energy transfer along the
SSM
frequency f120 Hz. The amplitude of the motion at Pis small
enough for that point to be considered a node.A node also exists at
Q. (a) What mass mallows the oscillator to set up the fourth har-
monic on the string? (b) What standing wave mode, if any, can be
set up if m1.00 kg?
•••59 In Fig. 16-43,
an aluminum wire, of
length L160.0 cm,
cross-sectional area 1.00
102cm2, and density
2.60 g/cm3, is joined to a
steel wire, of density
7.80 g/cm3and the same
cross-sectional area.The
compound wire, loaded with a block of mass m10.0 kg, is
arranged so that the distance L2from the joint to the supporting
pulley is 86.6 cm.Transverse waves are set up on the wire by an ex-
ternal source of variable frequency; a node is located at the pulley.
(a) Find the lowest frequency that generates a standing wave hav-
ing the joint as one of the nodes. (b) How many nodes are
observed at this frequency?
•••60 In Fig. 16-42, a string, tied to a sinusoidal oscillator at P
and running over a support at Q, is stretched by a block of mass m.
The separation Lbetween Pand Qis 1.20 m, and the frequency f
of the oscillator is fixed at 120 Hz. The amplitude of the motion at
Pis small enough for that point to be considered a node. A node
also exists at Q. A standing wave appears when the mass of the
hanging block is 286.1 g or 447.0 g, but not for any intermediate
mass.What is the linear density of the string?
Additional Problems
61 In an experiment on standing waves, a string 90 cm long is
attached to the prong of an electrically driven tuning fork that os-
cillates perpendicular to the length of the string at a frequency of
60 Hz. The mass of the string is 0.044 kg. What tension must the
string be under (weights are attached to the other end) if it is to os-
cillate in four loops?
62 A sinusoidal transverse wave traveling in the positive
direction of an xaxis has an amplitude of 2.0 cm, a wavelength of
10 cm, and a frequency of 400 Hz. If the wave equation is of the
form y(x,t)ymsin(kx vt), what are (a) ym, (b) k, (c) v, and
(d) the correct choice of sign in front of v? What are (e) the maxi-
mum transverse speed of a point on the cord and (f) the speed of
the wave?
63 A wave has a speed of 240 m/s and a wavelength of 3.2 m.What
are the (a) frequency and (b) period of the wave?
64 The equation of a transverse wave traveling along a string is
y0.15 sin(0.79x13t),
in which xand yare in meters and tis in seconds. (a) What is the dis-
placement yat x2.3 m, t0.16 s? A second wave is to be added
to the first wave to produce standing waves on the string. If the sec-
ond wave is of the form y(x,t)ymsin(kx vt),what are (b) ym, (c)
k, (d) v, and (e) the correct choice of sign in front of vfor this sec-
ond wave? (f) What is the displacement of the resultant standing
wave at x2.3 m, t0.16 s?
65 The equation of a transverse wave traveling along a string is
y(2.0 mm) sin[(20 m1)x(600 s1)t].
Find the (a) amplitude, (b) frequency, (c) velocity (including
Figure 16-44 Problem 66.
ys
–ys
y (mm)
t
Figure 16-45 Problem 68.
hs
123450
h(x)
t= 0
x
0
Figure 16-46 Problem 70.
as
–as
ay (m/s2)
t
m
L1L2
Aluminum Steel
Figure 16-43 Problem 59.
477
PROBLEMS
string? (g) What is the transverse displacement ywhen this mini-
mum transfer occurs?
72 Two sinusoidal 120 Hz
waves, of the same frequency
and amplitude, are to be sent in
the positive direction of an xaxis
that is directed along a cord un-
der tension. The waves can be
sent in phase, or they can be
phase-shifted. Figure 16-47
shows the amplitude yof the re-
sulting wave versus the distance of the shift (how far one wave is
shifted from the other wave). The scale of the vertical axis is set
by y
s6.0 mm. If the equations for the two waves are of the
form y(x,t)ymsin(kx vt), what are (a) ym, (b) k, (c) v, and
(d) the correct choice of sign in front of v?
73 At time t0 and at position x0 m along a string, a travel-
ing sinusoidal wave with an angular frequency of 440 rad/s has dis-
placement y4.5 mm and transverse velocity u0.75 m/s. If
the wave has the general form y(x,t)ymsin(kx vtf), what
is phase constant f?
74 Energy is transmitted at rate P1by a wave of frequency f1on a
string under tension t1.What is the new energy transmission rate P2
in terms of P1(a) if the tension is increased to t24t1and (b) if, in-
stead,the frequency is decreased to f2f1/2?
75 (a) What is the fastest transverse wave that can be sent along
a steel wire? For safety reasons, the maximum tensile stress to
which steel wires should be subjected is 7.00 108N/m2. The den-
sity of steel is 7800 kg/m3. (b) Does your answer depend on the di-
ameter of the wire?
76 A standing wave results from the sum of two transverse trav-
eling waves given by
y10.050 cos(px4pt)
and y20.050 cos(px4pt),
where x,y1, and y2are in meters and tis in seconds. (a) What is the
smallest positive value of xthat corresponds to a node? Beginning
at t0, what is the value of the (b) first, (c) second, and (d) third
time the particle at x0 has zero velocity?
77 The type of rubber band used inside some baseballs
and golf balls obeys Hooke’s law over a wide range of elonga-
tion of the band. A segment of this material has an unstretched
length and a mass m. When a force Fis applied, the band
stretches an additional length . (a) What is the speed (in
terms of m,, and the spring constant k) of transverse waves
on this stretched rubber band? (b) Using your answer to (a),
show that the time required for a transverse pulse to travel the
length of the rubber band is proportional to if
and is constant if .
78 The speed of electromagnetic waves (which include visible
light, radio, and x rays) in vacuum is 3.0 108m/s. (a) Wavelengths
of visible light waves range from about 400 nm in the violet to
about 700 nm in the red.What is the range of frequencies of these
waves? (b) The range of frequencies for shortwave radio (for ex-
ample, FM radio and VHF television) is 1.5 to 300 MHz. What is
the corresponding wavelength range? (c) X-ray wavelengths range
from about 5.0 nm to about 1.0 102nm. What is the frequency
range for x rays?
1/1
SSM
79 A 1.50 m wire has a mass of 8.70 g and is under a ten-
sion of 120 N. The wire is held rigidly at both ends and set into
oscillation. (a) What is the speed of waves on the wire? What is the
wavelength of the waves that produce (b) one-loop and (c) two-
loop standing waves? What is the frequency of the waves that pro-
duce (d) one-loop and (e) two-loop standing waves?
80 When played in a certain manner, the lowest resonant fre-
quency of a certain violin string is concert A (440 Hz). What is the
frequency of the (a) second and (b) third harmonic of the string?
81 A sinusoidal transverse wave traveling in the negative
direction of an xaxis has an amplitude of 1.00 cm, a frequency of
550 Hz, and a speed of 330 m/s. If the wave equation is of the form
y(x,t)ymsin(kx vt), what are (a) ym, (b) v, (c) k, and (d) the
correct choice of sign in front of v?
82 Two sinusoidal waves of the same wavelength travel in the same
direction along a stretched string. For wave 1, ym3.0 mm and f
0; for wave 2, ym5.0 mm and f70. What are the (a) amplitude
and (b) phase constant of the resultant wave?
83 A sinusoidal transverse wave of amplitude ymand
wavelength ltravels on a stretched cord. (a) Find the ratio of
the maximum particle speed (the speed with which a single particle
in the cord moves transverse to the wave) to the wave speed. (b)
Does this ratio depend on the material of which the cord is made?
84 Oscillation of a 600 Hz tuning fork sets up standing waves in a
string clamped at both ends. The wave speed for the string is
400 m/s. The standing wave has four loops and an amplitude of
2.0 mm. (a) What is the length of the string? (b) Write an equation
for the displacement of the string as a function of position and time.
85 A 120 cm length of string is stretched between fixed supports.
What are the (a) longest, (b) second longest, and (c) third longest
wavelength for waves traveling on the string if standing waves are to
be set up? (d) Sketch those standing waves.
86 (a) Write an equation describing a sinusoidal transverse wave
traveling on a cord in the positive direction of a yaxis with an an-
gular wave number of 60 cm1, a period of 0.20 s, and an amplitude
of 3.0 mm. Take the transverse direction to be the zdirection.
(b) What is the maximum transverse speed of a point on the cord?
87 A wave on a string is described by
y(x,t)15.0 sin(px/8 4pt),
where xand yare in centimeters and tis in seconds. (a) What is
the transverse speed for a point on the string at x6.00 cm
when t0.250 s? (b) What is the maximum transverse speed of
any point on the string? (c) What is the magnitude of the
transverse acceleration for a point on the string at x6.00 cm
when t0.250 s? (d) What is the magnitude of the maximum
transverse acceleration for any point on the string?
88 Body armor. When a high-speed projectile such as a
bullet or bomb fragment strikes modern body armor, the fabric of
the armor stops the projectile and prevents penetration by quickly
spreading the projectile’s energy over a large area. This spreading
is done by longitudinal and transverse pulses that move radially
from the impact point, where the projectile pushes a cone-shaped
dent into the fabric. The longitudinal pulse, racing along the fibers
of the fabric at speed vlahead of the denting, causes the fibers to
thin and stretch, with material flowing radially inward into the
dent. One such radial fiber is shown in Fig. 16-48a. Part of the pro-
jectile’s energy goes into this motion and stretching.The transverse
SSM
SSM
Figure 16-47 Problem 72.
y' (mm)
–y'
s
0
y'
s
10
Shift distance (cm)
20
478 CHAPTER 16 WAVES—I
pulse, moving at a slower speed vt, is due to the denting. As the
projectile increases the dent’s depth, the dent increases in radius,
causing the material in the fibers to move in the same direction as the
projectile (perpendicular to the transverse pulse’s direction of
travel). The rest of the projectile’s energy goes into this motion. All
the energy that does not eventually go into permanently deforming
the fibers ends up as thermal energy.
Figure 16-48bis a graph of speed vversus time tfor a bullet of
mass 10.2 g fired from a .38 Special revolver directly into body ar-
mor. The scales of the vertical and horizontal axes are set by vs
300 m/s and ts40.0 ms. Take vl2000 m/s, and assume that the
half-angle uof the conical dent is 60. At the end of the collision,
what are the radii of (a) the thinned region and (b) the dent (as-
suming that the person wearing the armor remains stationary)?
moving upward in the positive direction of a yaxis with a trans-
verse velocity of 5.0 m/s. What are (a) the amplitude of the mo-
tion of that point and (b) the tension in the rope? (c) Write the
standing wave equation for the fundamental mode.
92 Two waves,
y1(2.50 mm) sin[(25.1 rad/m)x(440 rad/s)t]
and y2(1.50 mm) sin[(25.1 rad/m)x(440 rad/s)t],
travel along a stretched string. (a) Plot the resultant wave as
a function of tfor x0, l/8, l/4, 3l/8, and l/2, where lis the
wavelength. The graphs should extend from t0 to a little over
one period. (b) The resultant wave is the superposition of a stand-
ing wave and a traveling wave. In which direction does the travel-
ing wave move? (c) How can you change the original waves so
Radius reached
by longitudinal
pulse
Radius reached by
transverse pulse
v
θ
Bullet
vlvtvtvl
(a)
(b)
vs
0
t ( s)
ts
v (m/s)
μ
Figure 16-48 Problem 88.
89 Two waves are described by
y10.30 sin[p (5x200t)]
and y20.30 sin[p (5x200t)p/3],
where y1,y2, and xare in meters and tis in seconds. When these two
waves are combined, a traveling wave is produced. What are the (a)
amplitude, (b) wave speed, and (c) wavelength of that traveling wave?
90 A certain transverse sinu-
soidal wave of wavelength 20 cm
is moving in the positive direc-
tion of an xaxis. The transverse
velocity of the particle at x0
as a function of time is shown in
Fig. 16-49, where the scale of
the vertical axis is set by us5.0 cm/s. What are the (a) wave
speed, (b) amplitude, and (c) frequency? (d) Sketch the wave
between x0 and x20 cm at t2.0 s.
91 In a demonstration, a 1.2 kg horizontal rope is fixed in
SSM
the resultant wave is the superposition of standing and traveling
waves with the same amplitudes as before but with the traveling
wave moving in the opposite direction? Next, use your graphs to
find the place at which the oscillation amplitude is (d) maximum
and (e) minimum. (f) How is the maximum amplitude related to
the amplitudes of the original two waves? (g) How is the minimum
amplitude related to the amplitudes of the original two waves?
93 A traveling wave on a string is described by
where xand yare in centimeters and tis in seconds.(a) For t0, plot y
as a function of xfor 0 x160 cm. (b) Repeat (a) for t0.05 s and
t0.10 s. From your graphs, determine (c) the wave speed and (d) the
direction in which the wave is traveling.
94 In Fig. 16-50, a circular loop of string is set
spinning about the center point in a place with
negligible gravity. The radius is 4.00 cm and the
tangential speed of a string segment is 5.00
cm/s. The string is plucked. At what speed do
transverse waves move along the string? (Hint:
Apply Newton’s second law to a small, but fi-
nite, section of the string.)
95 A continuous traveling wave with amplitude Ais incident on
a boundary.The continuous reflection, with a smaller amplitude B,
travels back through the incoming wave.The resulting interference
pattern is displayed in Fig. 16-51. The standing wave ratio is
defined to be
SWR AB
AB.
y2.0 sin2p
t
0.40 x
80
,
Figure 16-50
Problem 94.
Figure 16-49 Problem 90.
1 2 34
us
–us
0
u(cm/s)
t(s)
place at its two ends (x0 and x2.0 m) and made to oscil-
late up and down in the fundamental mode, at frequency 5.0 Hz.
At t0, the point at x1.0 m has zero displacement and is
A
max
A
min
A
max
Figure 16-51 Problem 95.
The reflection coefficient R
is the ratio of the power of
the reflected wave to the
power of the incoming wave
and is thus proportional to
the ratio (B/A)2.What is the
SWR for (a) total reflection
and (b) no reflection? (c) For SWR 1.50, what is Rexpressed as a
percentage?
96 Consider a loop in the standing wave created by two
waves (amplitude 5.00 mm and frequency 120 Hz) traveling in op-
posite directions along a string with length 2.25 m and mass 125 g
and under tension 40 N. At what rate does energy enter the loop
from (a) each side and (b) both sides? (c) What is the maximum ki-
netic energy of the string in the loop during its oscillation?
(speed of sound).
In air at 20C, the speed of sound is 343 m/s.
vAB
479
What Is Physics?
The physics of sound waves is the basis of countless studies in the research
journals of many fields. Here are just a few examples. Some physiologists are
concerned with how speech is produced, how speech impairment might be
corrected, how hearing loss can be alleviated, and even how snoring is pro-
duced. Some acoustic engineers are concerned with improving the acoustics of
cathedrals and concert halls, with reducing noise near freeways and road
construction, and with reproducing music by speaker systems. Some aviation en-
gineers are concerned with the shock waves produced by supersonic aircraft and
the aircraft noise produced in communities near an airport. Some medical re-
searchers are concerned with how noises produced by the heart and lungs can
signal a medical problem in a patient. Some paleontologists are concerned with
how a dinosaur’s fossil might reveal the dinosaur’s vocalizations. Some military
engineers are concerned with how the sounds of sniper fire might allow a sol-
dier to pinpoint the sniper’s location, and, on the gentler side, some biolo-
gists are concerned with how a cat purrs.
To begin our discussion of the physics of sound, we must first answer the
question “What are sound waves?”
Sound Waves
As we saw in Chapter 16, mechanical waves are waves that require a material
medium to exist. There are two types of mechanical waves: Transverse waves
involve oscillations perpendicular to the direction in which the wave travels;
longitudinal waves involve oscillations parallel to the direction of wave travel.
In this book, a sound wave is defined roughly as any longitudinal wave.
Seismic prospecting teams use such waves to probe Earth’s crust for oil. Ships
CHAPTER 17
Waves—II
17-1 SPEED OF SOUND
After reading this module, you should be able to . . .
17.01 Distinguish between a longitudinal wave and a
transverse wave.
17.02 Explain wavefronts and rays.
17.03 Apply the relationship between the speed of sound
through a material, the material’s bulk modulus, and the
material’s density.
Key Idea
Learning Objectives
●Sound waves are longitudinal mechanical waves that can
travel through solids, liquids, or gases. The speed vof a
sound wave in a medium having bulk modulus Band den-
sity ris
17.04 Apply the relationship between the speed of sound, the
distance traveled by a sound wave, and the time required
to travel that distance.
480 CHAPTER 17 WAVES—II
carry sound-ranging gear (sonar) to detect underwater obstacles. Submarines
use sound waves to stalk other submarines, largely by listening for the charac-
teristic noises produced by the propulsion system. Figure 17-1 suggests how
sound waves can be used to explore the soft tissues of an animal or human body.
In this chapter we shall focus on sound waves that travel through the air and that
are audible to people.
Figure 17-2 illustrates several ideas that we shall use in our discussions.
Point Srepresents a tiny sound source, called a point source, that emits sound
waves in all directions. The wavefronts and rays indicate the direction of travel
and the spread of the sound waves. Wavefronts are surfaces over which the
oscillations due to the sound wave have the same value; such surfaces are rep-
resented by whole or partial circles in a two-dimensional drawing for a point
source. Rays are directed lines perpendicular to the wavefronts that indicate
the direction of travel of the wavefronts. The short double arrows superim-
posed on the rays of Fig. 17-2 indicate that the longitudinal oscillations of the
air are parallel to the rays.
Near a point source like that of Fig. 17-2, the wavefronts are spherical and
spread out in three dimensions, and there the waves are said to be spherical. As
the wavefronts move outward and their radii become larger, their curvature
decreases. Far from the source, we approximate the wavefronts as planes (or lines
on two-dimensional drawings), and the waves are said to be planar.
The Speed of Sound
The speed of any mechanical wave, transverse or longitudinal, depends on both an
inertial property of the medium (to store kinetic energy) and an elastic property of
the medium (to store potential energy). Thus, we can generalize Eq. 16-26, which
gives the speed of a transverse wave along a stretched string,by writing
(17-1)
where (for transverse waves) tis the tension in the string and mis the string’s
linear density. If the medium is air and the wave is longitudinal, we can guess
that the inertial property, corresponding to m, is the volume density rof air.
What shall we put for the elastic property?
In a stretched string, potential energy is associated with the periodic stretching
of the string elements as the wave passes through them. As a sound wave passes
through air, potential energy is associated with periodic compressions and expan-
sions of small volume elements of the air.The property that determines the extent
to which an element of a medium changes in volume when the pressure (force per
unit area) on it changes is the bulk modulus B,defined (from Eq. 12-25) as
(definition of bulk modulus). (17-2)
Here V/Vis the fractional change in volume produced by a change in pressure
p. As explained in Module 14-1, the SI unit for pressure is the newton per
square meter,which is given a special name, the pascal (Pa). From Eq. 17-2 we see
that the unit for Bis also the pascal. The signs of pand Vare always
opposite: When we increase the pressure on an element (pis positive), its vol-
ume decreases (Vis negative). We include a minus sign in Eq. 17-2 so that Bis
always a positive quantity. Now substituting Bfor tand rfor min Eq. 17-1 yields
(speed of sound) (17-3)vAB
B p
V/V
vAt
mAelastic property
inertial property ,
Mauro Fermariello/SPL/Photo Researchers, Inc.
Figure 17-1 A loggerhead turtle is being
checked with ultrasound (which has a
frequency above your hearing range); an
image of its interior is being produced on
a monitor off to the right.
Ray
Ray
S
Wavefronts
Figure 17-2 A sound wave travels from a
point source Sthrough a three-dimen-
sional medium. The wavefronts form
spheres centered on S; the rays are radi-
al to S. The short, double-headed arrows
indicate that elements of the medium
oscillate parallel to the rays.
481
17-1 SPEED OF SOUND
as the speed of sound in a medium with bulk modulus Band density r.Table 17-1
lists the speed of sound in various media.
The density of water is almost 1000 times greater than the density of air. If this
were the only relevant factor, we would expect from Eq. 17-3 that the speed of
sound in water would be considerably less than the speed of sound in air. However,
Table 17-1 shows us that the reverse is true.We conclude (again from Eq. 17-3) that
the bulk modulus of water must be more than 1000 times greater than that of air.
This is indeed the case.Water is much more incompressible than air, which (see Eq.
17-2) is another way of saying that its bulk modulus is much greater.
Formal Derivation of Eq. 17-3
We now derive Eq. 17-3 by direct application of Newton’s laws. Let a single
pulse in which air is compressed travel (from right to left) with speed vthrough the
air in a long tube, like that in Fig.16-2. Let us run along with the pulse at that speed,
so that the pulse appears to stand still in our reference frame. Figure 17-3a
shows the situation as it is viewed from that frame. The pulse is standing still, and
air is moving at speed vthrough it from left to right.
Let the pressure of the undisturbed air be pand the pressure inside the
pulse be pp, where pis positive due to the compression. Consider an element
of air of thickness xand face area A, moving toward the pulse at speed v.As this
element enters the pulse, the leading face of the element encounters a region of
higher pressure, which slows the element to speed vv, in which vis negative.
This slowing is complete when the rear face of the element reaches the pulse, which
requires time interval
(17-4)
Let us apply Newton’s second law to the element. During t, the average
force on the element’s trailing face is pA toward the right, and the average force
on the leading face is ( pp)Atoward the left (Fig. 17-3b). Therefore, the
average net force on the element during tis
FpA (pp)A
p A (net force). (17-5)
The minus sign indicates that the net force on the air element is directed to the
left in Fig. 17-3b. The volume of the element is Ax, so with the aid of Eq. 17-4,
we can write its mass as
mrVrAxrAv t(mass). (17-6)
The average acceleration of the element during tis
(acceleration). (17-7)av
t
tx
v.
Table 17-1 The Speed of Sounda
Medium Speed (m/s)
Gases
Air (0C) 331
Air (20C) 343
Helium 965
Hydrogen 1284
Liquids
Water (0C) 1402
Water (20C) 1482
Seawaterb1522
Solids
Aluminum 6420
Steel 5941
Granite 6000
aAt 0C and 1 atm pressure, except where noted.
bAt 20C and 3.5% salinity.
Δ
Δx
(b)
pA (p + p)A
Pulse
Moving air (fluid element)
p,v
p + p,v + vΔΔ
Δx
A
p,v
(a)
v
Figure 17-3 A compression pulse is sent from right to left down a long air-filled tube.The refer-
ence frame of the figure is chosen so that the pulse is at rest and the air moves from left to
right. (a) An element of air of width xmoves toward the pulse with speed v.(b) The leading
face of the element enters the pulse. The forces acting on the leading and trailing faces (due to
air pressure) are shown.
482 CHAPTER 17 WAVES—II
17-2 TRAVELING SOUND WAVES
After reading this module, you should be able to . . .
17.05 For any particular time and position, calculate the dis-
placement s(x,t)of an element of air as a sound wave travels
through its location.
17.06 Given a displacement function s(x,t)for a sound wave,
calculate the time between two given displacements.
17.07 Apply the relationships between wave speed v, angular
frequency v, angular wave number k, wavelength l, period
T, and frequency f.
17.08 Sketch a graph of the displacement s(x)of an element
of air as a function of position, and identify the amplitude
smand wavelength l.
17.09 For any particular time and position, calculate the pres-
sure variation (variation from atmospheric pressure) of an
element of air as a sound wave travels through its location.
17.10 Sketch a graph of the pressure variation p(x)of an el-
ement as a function of position, and identify the amplitude
pmand wavelength l.
17.11 Apply the relationship between pressure-variation am-
plitude pmand displacement amplitude sm.
17.12 Given a graph of position sversus time for a sound
wave, determine the amplitude smand the period T.
17.13 Given a graph of pressure variation pversus time
for a sound wave, determine the amplitude pmand the
period T.
p
Learning Objectives
●A sound wave causes a longitudinal displacement sof a
mass element in a medium as given by
ssmcos(kx vt),
where smis the displacement amplitude (maximum displace-
ment) from equilibrium, k2p/l, and v2pf,land fbeing
the wavelength and frequency, respectively, of the sound wave.
●The sound wave also causes a pressure change of the
medium from the equilibrium pressure:
ppmsin(kx vt),
where the pressure amplitude is
pm(vrv)sm.
p
Key Ideas
Thus, from Newton’s second law (Fma), we have, from Eqs. 17-5, 17-6,
and 17-7,
(17-8)
which we can write as
(17-9)
The air that occupies a volume V(Av t) outside the pulse is compressed by an
amount V(Avt) as it enters the pulse.Thus,
(17-10)
Substituting Eq. 17-10 and then Eq. 17-2 into Eq. 17-9 leads to
(17-11)
Solving for vyields Eq. 17-3 for the speed of the air toward the right in Fig. 17-3,
and thus for the actual speed of the pulse toward the left.
v2 p
v/v p
V/VB.
V
VAvt
Av tv
v.
v2 p
v/v.
pA(
Av t)v
t,
Traveling Sound Waves
Here we examine the displacements and pressure variations associated with a
sinusoidal sound wave traveling through air. Figure 17-4adisplays such a wave
traveling rightward through a long air-filled tube. Recall from Chapter 16 that
we can produce such a wave by sinusoidally moving a piston at the left end of
483
17-2 TRAVELING SOUND WAVES
the tube (as in Fig. 16-2). The piston’s rightward motion moves the element of
air next to the piston face and compresses that air; the piston’s leftward motion
allows the element of air to move back to the left and the pressure to decrease.
As each element of air pushes on the next element in turn, the right–left motion
of the air and the change in its pressure travel along the tube as a sound wave.
Consider the thin element of air of thickness xshown in Fig. 17-4b. As the
wave travels through this portion of the tube, the element of air oscillates left
and right in simple harmonic motion about its equilibrium position. Thus, the
oscillations of each air element due to the traveling sound wave are like those of
a string element due to a transverse wave, except that the air element oscillates
longitudinally rather than transversely. Because string elements oscillate parallel
to the yaxis, we write their displacements in the form y(x,t). Similarly, because
air elements oscillate parallel to the xaxis, we could write their displacements in
the confusing form x(x,t), but we shall use s(x,t) instead.
Displacement. To show that the displacements s(x,t) are sinusoidal func-
tions of xand t, we can use either a sine function or a cosine function. In this
chapter we use a cosine function, writing
s(x,t)smcos(kx vt). (17-12)
Figure 17-5alabels the various parts of this equation. In it, smis the displacement
amplitude —that is, the maximum displacement of the air element to either side
of its equilibrium position (see Fig. 17-4b). The angular wave number k, angular
frequency v, frequency f, wavelength l, speed v, and period Tfor a sound
(longitudinal) wave are defined and interrelated exactly as for a transverse
wave, except that lis now the distance (again along the direction of travel) in
which the pattern of compression and expansion due to the wave begins to
repeat itself (see Fig.17-4a). (We assume smis much less than l.)
Pressure. As the wave moves, the air pressure at any position xin Fig. 17-4a
varies sinusoidally, as we prove next.To describe this variation we write
p(x,t)pmsin(kx vt). (17-13)
Figure 17-5blabels the various parts of this equation. A negative value of pin
Eq. 17-13 corresponds to an expansion of the air, and a positive value to a com-
pression. Here pmis the pressure amplitude, which is the maximum increase or
decrease in pressure due to the wave; pmis normally very much less than the
pressure ppresent when there is no wave. As we shall prove, the pressure ampli-
C
ompress
i
on
(a)
Δ x
Expansion
λ
Equilibrium
position
s
smsm
(b)
x
Oscillating fluid element
v
The element oscillates
left and right as the wave
moves through it.
Figure 17-4 (a) A sound wave, traveling
through a long air-filled tube with speed v,
consists of a moving, periodic pattern of
expansions and compressions of the air.
The wave is shown at an arbitrary instant.
(b) A horizontally expanded view of a
short piece of the tube. As the wave pass-
es, an air element of thickness xoscillates
left and right in simple harmonic motion
about its equilibrium position. At the
instant shown in (b), the element happens
to be displaced a distance sto the right of
its equilibrium position. Its maximum dis-
placement, either right or left, is sm.
Displacement
Pressure variation
Pressure amplitude
Displacement
amplitude
ω
ω
Oscillating
term
(a)
ω
ω
(b)
s(x,t) = sm cos(kx – t)
Δp(x,t) = Δpm sin(kx – t)
Figure 17-5 (a) The displacement function
and (b) the pressure-variation function
of a traveling sound wave consist of an
amplitude and an oscillating term.
484 CHAPTER 17 WAVES—II
tude pmis related to the displacement amplitude smin Eq. 17-12 by
pm(vrv)sm. (17-14)
Figure 17-6 shows plots of Eqs. 17-12 and 17-13 at t0; with time, the
two curves would move rightward along the horizontal axes. Note that the
displacement and pressure variation are p/2 rad (or 90) out of phase. Thus, for
example, the pressure variation pat any point along the wave is zero when the
displacement there is a maximum.
Checkpoint 1
When the oscillating air element in Fig.17-4bis moving rightward through the point
of zero displacement, is the pressure in the element at its equilibrium value, just be-
ginning to increase, or just beginning to decrease?
Figure 17-6 (a) A plot of the displacement
function (Eq. 17-12) for t0. (b) A simi-
lar plot of the pressure-variation function
(Eq. 17-13). Both plots are for a 1000 Hz
sound wave whose pressure amplitude is
at the threshold of pain.
20 40 60 80
20 40 60 80
10
0
–10
x(cm)
t = 0
(a)
sm
pm
Δ
x(cm)
t = 0
Pressure variation (Pa) Displacement ( m)
μ
30
20
10
0
–10
–20
–30
(b)
Derivation of Eqs. 17-13 and 17-14
Figure 17-4bshows an oscillating element of air of cross-sectional area Aand
thickness x, with its center displaced from its equilibrium position by
distance s. From Eq. 17-2 we can write, for the pressure variation in the dis-
placed element,
(17-15)
The quantity Vin Eq. 17-15 is the volume of the element, given by
VAx. (17-16)
The quantity Vin Eq. 17-15 is the change in volume that occurs when the
element is displaced. This volume change comes about because the displace-
ments of the two faces of the element are not quite the same, differing by some
amount s.Thus, we can write the change in volume as
VAs. (17-17)
Substituting Eqs. 17-16 and 17-17 into Eq. 17-15 and passing to the differen-
tial limit yield
(17-18)
The symbols indicate that the derivative in Eq. 17-18 is a partial derivative,
which tells us how schanges with xwhen the time tis fixed. From Eq. 17-12 we
then have, treating tas a constant,
Substituting this quantity for the partial derivative in Eq. 17-18 yields
pBksmsin(kx vt).
This tells us that the pressure varies as a sinusoidal function of time and that the
amplitude of the variation is equal to the terms in front of the sine function.
Setting pmBksm, this yields Eq. 17-13, which we set out to prove.
Using Eq. 17-3, we can now write
pm(Bk)sm(v2rk)sm.
Equation 17-14, which we also wanted to prove, follows at once if we substitute
v/vfor kfrom Eq. 16-12.
s
x
x [smcos(kx vt)] ksmsin(kx vt).
pBs
xBs
x.
pBV
V.
485
17-3 INTERFERENCE
1.1 105m11 mm. (Answer)
That is only about one-seventh the thickness of a book page.
Obviously, the displacement amplitude of even the loudest
sound that the ear can tolerate is very small. Temporary ex-
posure to such loud sound produces temporary hearing loss,
probably due to a decrease in blood supply to the inner ear.
Prolonged exposure produces permanent damage.
The pressure amplitude pmfor the faintest detectable
sound at 1000 Hz is 2.8 105Pa. Proceeding as above
leads to sm1.1 1011 m or 11 pm, which is about one-
tenth the radius of a typical atom. The ear is indeed a sensi-
tive detector of sound waves.
sm28 Pa
(343 m/s)(1.21 kg/m3)(2
)(1000 Hz)
Sample Problem 17.01 Pressure amplitude, displacement amplitude
The maximum pressure amplitude pmthat the human ear can
tolerate in loud sounds is about 28 Pa (which is very much less
than the normal air pressure of about 105Pa). What is the dis-
placement amplitude smfor such a sound in air of density r
1.21 kg/m3,at a frequency of 1000 Hz and a speed of 343 m/s?
KEY IDEA
The displacement amplitude smof a sound wave is related to
the pressure amplitude pmof the wave according to Eq.17-14.
Calculations: Solving that equation for smyields
.
Substituting known data then gives us
smpm
v
pm
v
(2
f)
Additional examples, video, and practice available at WileyPLUS
17-3 INTERFERENCE
Learning Objectives
waves with the same amplitude, wavelength, and travel di-
rection, determine the type of interference between the
waves (fully destructive interference, fully constructive in-
terference, or indeterminate interference).
17.16 Convert a phase difference between radians, degrees,
and number of wavelengths.
●The interference of two sound waves with identical wave-
lengths passing through a common point depends on their phase
difference fthere. If the sound waves were emitted in phase and
are traveling in approximately the same direction, fis given by
where Lis their path length difference.
●Fully constructive interference occurs when fis an integer
multiple of 2p,
fL
l 2p,
fm(2p), for m0,1,2,...,
and, equivalently, when Lis related to wavelength lby
0,1,2,....
●Fully destructive interference occurs when fis an odd multiple
of p,f(2m1)p, for m0,1,2,...,
0.5, 1.5, 2.5,....
L
l
L
l
After reading this module, you should be able to . . .
17.14 If two waves with the same wavelength begin in
phase but reach a common point by traveling along dif-
ferent paths, calculate their phase difference fat that
point by relating the path length difference Lto the
wavelength l.
17.15 Given the phase difference between two sound
Key Ideas
Interference
Like transverse waves, sound waves can undergo interference. In fact, we can
write equations for the interference as we did in Module 16-5 for transverse
waves. Suppose two sound waves with the same amplitude and wavelength are
traveling in the positive direction of an xaxis with a phase difference of f.We can
express the waves in the form of Eqs. 16-47 and 16-48 but, to be consistent with
Eq. 17-12, we use cosine functions instead of sine functions:
s1(x,t)sm cos(kx vt)
and
486 CHAPTER 17 WAVES—II
S1
L1
L2
S2
P
P
P
(a)
(b)
(c)
The interference at P
depends on the difference
in the path lengths to reach P.
If the difference is equal to,
say, 2.0 , then the waves
arrive exactly in phase. This
is how transverse waves
would look.
λ
If the difference is equal to,
say, 2.5 , then the waves
arrive exactly out of phase.
This is how transverse
waves would look.
λ
Figure 17-7 (a) Two point sources S1and S2
emit spherical sound waves in phase. The
rays indicate that the waves pass through a
common point P. The waves (represented
with transverse waves) arrive at P(b) exactly
in phase and (c) exactly out of phase.
and
These waves overlap and interfere. From Eq. 16-51, we can write the resultant
wave as
As we saw with transverse waves, the resultant wave is itself a traveling wave. Its
amplitude is the magnitude
(17-19)
As with transverse waves, the value of fdetermines what type of interference the
individual waves undergo.
One way to control fis to send the waves along paths with different lengths.
Figure 17-7ashows how we can set up such a situation: Two point sources S1and
S2emit sound waves that are in phase and of identical wavelength l. Thus, the
sources themselves are said to be in phase; that is, as the waves emerge from the
sources, their displacements are always identical.We are interested in the waves
that then travel through point Pin Fig. 17-7a.We assume that the distance to P
is much greater than the distance between the sources so that we can approxi-
mate the waves as traveling in the same direction at P.
If the waves traveled along paths with identical lengths to reach point P,
they would be in phase there. As with transverse waves, this means that they
would undergo fully constructive interference there. However, in Fig. 17-7a, path
L2traveled by the wave from S2is longer than path L1traveled by the wave from
S1. The difference in path lengths means that the waves may not be in
phase at point P. In other words, their phase difference fat Pdepends on their
path length difference L|L2L1|.
To relate phase difference fto path length difference L, we recall (from
Module 16-1) that a phase difference of 2prad corresponds to one wavelength.
Thus, we can write the proportion
, (17-20)
from which
(17-21)
Fully constructive interference occurs when fis zero, 2p, or any integer multiple
of 2p. We can write this condition as
fm(2p), for m0,1,2,... (fully constructive interference). (17-22)
From Eq. 17-21, this occurs when the ratio L/lis
0,1,2,... (fully constructive interference). (17-23)
For example, if the path length difference L|L2L1| in Fig. 17-7ais equal
to 2l, then L/l2 and the waves undergo fully constructive interference at
point P(Fig. 17-7b). The interference is fully constructive because the wave
from S2is phase-shifted relative to the wave from S1by 2l, putting the two
waves exactly in phase at P.
Fully destructive interference occurs when fis an odd multiple of p:
f(2m1)p, for m0,1,2,... (fully destructive interference). (17-24)
L
l
fL
l
2p.
f
2pL
l
s
m2sm cos1
2f.
s[2sm cos1
2f] cos(kx vt1
2f).
s2(x,t)sm cos(kx vtf).
487
17-3 INTERFERENCE
From Eq. 17-21, this occurs when the ratio L/lis
0.5, 1.5, 2.5,... (fully destructive interference). (17-25)
For example, if the path length difference L|L2L1| in Fig. 17-7ais equal to
2.5l, then L/l2.5 and the waves undergo fully destructive interference at
point P(Fig. 17-7c). The interference is fully destructive because the wave from
S2is phase-shifted relative to the wave from S1by 2.5 wavelengths, which puts the
two waves exactly out of phase at P.
Of course, two waves could produce intermediate interference as, say, when
L/l1.2. This would be closer to fully constructive interference (L/l1.0)
than to fully destructive interference (L/l1.5).
L
l
Reasoning: The wave from S1travels the extra distance D
(1.5l) to reach P2.Thus, the path length difference is
L1.5l. (Answer)
From Eq. 17-25, this means that the waves are exactly out
of phase at P2and undergo fully destructive interference
there.
(c) Figure 17-8dshows a circle with a radius much greater
than D, centered on the midpoint between sources S1and S2.
What is the number of points Naround this circle at which
the interference is fully constructive? (That is, at how many
points do the waves arrive exactly in phase?)
Reasoning: Starting at point a, let’s move clockwise
along the circle to point d.As we move, path length differ-
ence Lincreases and so the type of interference changes.
From (a), we know that is L0lat point a. From (b),
we know that L1.5lat point d. Thus, there must be
Sample Problem 17.02 Interference points along a big circle
In Fig. 17-8a, two point sources S1and S2, which are in phase
and separated by distance D1.5l, emit identical sound
waves of wavelength l.
(a) What is the path length difference of the waves from S1
and S2at point P1, which lies on the perpendicular bisector
of distance D, at a distance greater than Dfrom the sources
(Fig. 17-8b)? (That is, what is the difference in the distance
from source S1to point P1and the distance from source S2
to P1?) What type of interference occurs at P1?
Reasoning: Because the waves travel identical distances to
reach P1, their path length difference is
L0. (Answer)
From Eq. 17-23, this means that the waves undergo fully
constructive interference at P1because they start in phase at
the sources and reach P1in phase.
(b) What are the path length difference and type of inter-
ference at point P2in Fig.17-8c?
A
D/2
D/2
S1L1
L2
S2
P1
(b)
S1
S2
P2
(c)
D
S1
S2
(a)
a0
1.5
λ
λ
S2
S1
d
(d)
D
The difference in these
path lengths equals 0.
The difference in these
path lengths is D,
which equals 1.5 .
Thus, the waves arrive exactly
in phase and undergo fully
constructive interference.
Thus, the waves arrive
exactly out of phase
and undergo fully
destructive interference.
λ
Figure 17-8 (a) Two point sources S1and S2, separated by distance D, emit spherical sound waves in phase. (b) The waves travel equal
distances to reach point P1.(c) Point P2is on the line extending through S1and S2.(d) We move around a large circle. (Figure continues)
488 CHAPTER 17 WAVES—II
17-4 INTENSITY AND SOUND LEVEL
After reading this module, you should be able to . . .
17.17 Calculate the sound intensity Iat a surface as the ratio of
the power Pto the surface area A.
17.18 Apply the relationship between the sound intensity Iand
the displacement amplitude smof the sound wave.
17.19 Identify an isotropic point source of sound.
17.20 For an isotropic point source, apply the relationship in-
volving the emitting power Ps, the distance rto a detector,
and the sound intensity Iat the detector.
17.21 Apply the relationship between the sound level b,
the sound intensity I, and the standard reference
intensity I0.
17.22 Evaluate a logarithm function (log) and an antilogarithm
function (log1).
17.23 Relate the change in a sound level to the change in sound
intensity.
●The intensity Iof a sound wave at a surface is the average
rate per unit area at which energy is transferred by the wave
through or onto the surface:
,
where Pis the time rate of energy transfer (power) of the
sound wave and Ais the area of the surface intercepting
the sound. The intensity Iis related to the displacement
amplitude smof the sound wave by
I1
2
vv2s2
m.
IP
A
●The intensity at a distance rfrom a point source that
emits sound waves of power Psequally in all directions
(isotropically) is
.
●The sound level bin decibels (dB) is defined as
,
where I0(1012 W/m2)is a reference intensity level to
which all intensities are compared. For every factor-of-10
increase in intensity, 10 dB is added to the sound level.
b(10 dB) log I
I0
IPs
4pr2
Learning Objectives
Key Ideas
one point between aand dat which Ll(Fig. 17-8e).
From Eq. 17-23, fully constructive interference occurs at
that point.Also, there can be no other point along the way
from point ato point dat which fully constructive inter-
ference occurs, because there is no other integer than 1
between 0 at point aand 1.5 at point d.
We can now use symmetry to locate other points of fully
constructive or destructive interference (Fig. 17-8f).
Symmetry about line cd gives us point b, at which L0l.
Also, there are three more points at which Ll. In all
(Fig. 17-8g) we have
N6. (Answer)
Additional examples, video, and practice available at WileyPLUS
Figure 17-8 (continued) (e) Another point of fully constructive interference. ( f) Using symmetry to determine other points. (g) The six points
of fully constructive interference.
The difference
in these path
lengths
equals 1.0 .
00
1.0
λ
λλ
1.0
λ
1.0
λ
1.0
λ
S2
S1
(e)
λ
0ba0
1.5
λ
1.5
λ
1.0
λ
λλ
1.0
λ
1.0
λ
1.0
λ
S2
S1
d
c
(f)(g)
1.0
λ
Thus, the waves arrive exactly in phase
and undergo fully constructive interference.
We find six points
of fully constructive
interference.
Maximum phase
difference
Maximum phase difference
Zero phase
difference
Zero
phase
difference
489
17-4 INTENSITY AND SOUND LEVEL
S
r
Figure 17-9 A point source Semits sound
waves uniformly in all directions. The
waves pass through an imaginary sphere
of radius rthat is centered on S.
Checkpoint 2
The figure indicates three small patches 1, 2, and 3 that
lie on the surfaces of two imaginary spheres; the spheres
are centered on an isotropic point source Sof sound.
The rates at which energy is transmitted through the
three patches by the sound waves are equal. Rank the
patches according to (a) the intensity of the sound on
them and (b) their area, greatest first.
S
3
1
2
Intensity and Sound Level
If you have ever tried to sleep while someone played loud music nearby, you are
well aware that there is more to sound than frequency, wavelength, and speed.
There is also intensity. The intensity Iof a sound wave at a surface is the average
rate per unit area at which energy is transferred by the wave through or onto the
surface.We can write this as
, (17-26)
where Pis the time rate of energy transfer (the power) of the sound wave and
Ais the area of the surface intercepting the sound. As we shall derive shortly,
the intensity Iis related to the displacement amplitude smof the sound
wave by
. (17-27)
Intensity can be measured on a detector. Loudness is a perception, something
that you sense. The two can differ because your perception depends on factors
such as the sensitivity of your hearing mechanism to various frequencies.
Variation of Intensity with Distance
How intensity varies with distance from a real sound source is often complex.
Some real sources (like loudspeakers) may transmit sound only in particular
directions, and the environment usually produces echoes (reflected sound
waves) that overlap the direct sound waves. In some situations, however, we
can ignore echoes and assume that the sound source is a point source that
emits the sound isotropically —that is, with equal intensity in all directions.
The wavefronts spreading from such an isotropic point source Sat a particular
instant are shown in Fig.17-9.
Let us assume that the mechanical energy of the sound waves is conserved
as they spread from this source. Let us also center an imaginary sphere of radius
ron the source, as shown in Fig. 17-9. All the energy emitted by the source
must pass through the surface of the sphere. Thus, the time rate at which energy
is transferred through the surface by the sound waves must equal the time rate
at which energy is emitted by the source (that is, the power Psof the source).
From Eq. 17-26, the intensity Iat the sphere must then be
, (17-28)
where 4pr2is the area of the sphere. Equation 17-28 tells us that the intensity of
sound from an isotropic point source decreases with the square of the distance r
from the source.
IP
s
4pr2
I1
2rvv2s2
m
IP
A
490 CHAPTER 17 WAVES—II
The Decibel Scale
The displacement amplitude at the human ear ranges from about 105m for
the loudest tolerable sound to about 1011 m for the faintest detectable sound,
a ratio of 106. From Eq. 17-27 we see that the intensity of a sound varies as the
square of its amplitude, so the ratio of intensities at these two limits of the hu-
man auditory system is 1012. Humans can hear over an enormous range of
intensities.
We deal with such an enormous range of values by using logarithms.
Consider the relation
ylog x,
in which xand yare variables. It is a property of this equation that if we multiply
xby 10, then yincreases by 1.To see this, we write
ylog(10x)log 10 log x1y.
Similarly, if we multiply xby 1012,yincreases by only 12.
Thus, instead of speaking of the intensity Iof a sound wave, it is much more
convenient to speak of its sound level b, defined as
(17-29)
Here dB is the abbreviation for decibel, the unit of sound level, a name that was
chosen to recognize the work of Alexander Graham Bell. I0in Eq. 17-29 is a
standard reference intensity (1012 W/m2), chosen because it is near the lower
limit of the human range of hearing. For II0, Eq. 17-29 gives b10 log 1 0,
so our standard reference level corresponds to zero decibels. Then bincreases
by 10 dB every time the sound intensity increases by an order of magnitude
(a factor of 10). Thus, b40 corresponds to an intensity that is 104times
the standard reference level. Table 17-2 lists the sound levels for a variety of
environments.
Derivation of Eq. 17-27
Consider, in Fig. 17-4a, a thin slice of air of thickness dx, area A, and mass dm,
oscillating back and forth as the sound wave of Eq. 17-12 passes through it. The
kinetic energy dK of the slice of air is
(17-30)
Here vsis not the speed of the wave but the speed of the oscillating element of air,
obtained from Eq. 17-12 as
Using this relation and putting dm rAdxallow us to rewrite Eq. 17-30 as
dK (rA dx)(vsm)2sin2(kx vt). (17-31)
Dividing Eq. 17-31 by dt gives the rate at which kinetic energy moves along with
the wave. As we saw in Chapter 16 for transverse waves, dx/dt is the wave speed
v, so we have
rAvv2s2
msin2(kx vt). (17-32)
dK
dt 1
2
1
2
vss
tvsm sin(kx vt).
dK 1
2dm v2
s.
b(10 dB) log I
I0
.
Sound can cause the wall of a drinking
glass to oscillate. If the sound produces a
standing wave of oscillations and if the
intensity of the sound is large enough, the
glass will shatter.
© Ben Rose
Table 17-2 Some Sound Levels (dB)
Hearing threshold 0
Rustle of leaves 10
Conversation 60
Rock concert 110
Pain threshold 120
Jet engine 130
491
17-4 INTENSITY AND SOUND LEVEL
The average rate at which kinetic energy is transported is
(17-33)
To obtain this equation, we have used the fact that the average value of the
square of a sine (or a cosine) function over one full oscillation is .
We assume that potential energy is carried along with the wave at this
same average rate.The wave intensity I, which is the average rate per unit area
at which energy of both kinds is transmitted by the wave, is then, from
Eq. 17-33,
which is Eq. 17-27, the equation we set out to derive.
I2(dK/dt)avg
A1
2rvv2s2
m,
1
2
1
4rAvv2s2
m.
dK
dt
avg
1
2rAvv2s2
m[sin2(kx vt)]avg
Calculations: Putting these ideas together and noting that the
area of the cylindrical surface is A2prL,we have
. (17-34)
This tells us that the intensity of the sound from a line
source decreases with distance r(and not with the square of
distance ras for a point source). Substituting the given data,
we find
(Answer)
(b) At what time rate Pdis sound energy intercepted by an
acoustic detector of area Ad2.0 cm2, aimed at the spark
and located a distance r12 m from the spark?
Calculations: We know that the intensity of sound at the
detector is the ratio of the energy transfer rate Pdthere to
the detector’s area Ad:
(17-35)
We can imagine that the detector lies on the cylindrical
surface of (a).Then the sound intensity at the detector is the
intensity I(21.2 W/m2) at the cylindrical surface. Solving
Eq. 17-35 for Pdgives us
Pd(21.2 W/m2)(2.0 104m2)4.2 mW. (Answer)
IPd
Ad
.
21.2 W/m221 W/m2.
I1.6 104W
2p(12 m)(10 m)
IP
AP
s
2prL
Sample Problem 17.03 Intensity change with distance, cylindrical sound wave
An electric spark jumps along a straight line of length
L10 m, emitting a pulse of sound that travels radially
outward from the spark. (The spark is said to be a line
source of sound.) The power of this acoustic emission is
Ps1.6 104W.
(a) What is the intensity Iof the sound when it reaches a dis-
tance r12 m from the spark?
KEY IDEAS
(1) Let us center an imaginary cylinder of radius r12 m
and length L10 m (open at both ends) on the spark, as
shown in Fig. 17-10. Then the intensity Iat the cylindrical
surface is the ratio P/A, where Pis the time rate at which
sound energy passes through the surface and Ais the sur-
face area. (2) We assume that the principle of conservation
of energy applies to the sound energy. This means that the
rate Pat which energy is transferred through the cylinder
must equal the rate Psat which energy is emitted by the
source.
Additional examples, video, and practice available at WileyPLUS
r
L
Spark
Figure 17-10 A spark along a straight line of length Lemits sound
waves radially outward. The waves pass through an imaginary
cylinder of radius rand length Lthat is centered on the spark.
492 CHAPTER 17 WAVES—II
sound level as bfbi20 dB, we find
We next take the antilog of the far left and far right sides of
this equation. (Although the antilog 102.0 can be evaluated
mentally, you could use a calculator by keying in 10ˆ-2.0 or
using the 10xkey.) We find
(Answer)
Thus, the earplug reduces the intensity of the sound waves
to 0.010 of their initial intensity (two orders of magnitude).
If
Ii
log1 (2.0) 0.010.
log If
Ii
bfbi
10 dB 20 dB
10 dB 2.0.
Sample Problem 17.04 Decibels, sound level, change in intensity
Many veteran rockers suffer from acute hearing damage
because of the high sound levels they endured for years.
Many rockers now wear special earplugs to protect their
hearing during performances (Fig. 17-11). If an earplug de-
creases the sound level of the sound waves by 20 dB, what is
the ratio of the final intensity Ifof the waves to their initial
intensity Ii?
KEY IDEA
For both the final and initial waves, the sound level bis related
to the intensity by the definition of sound level in Eq.17-29.
Calculations: For the final waves we have
,
and for the initial waves we have
.
The difference in the sound levels is
. (17-36)
Using the identity
we can rewrite Eq. 17-36 as
. (17-37)
Rearranging and then substituting the given decrease in
bfbi(10 dB) log If
Ii
log a
blog c
dlog ad
bc ,
bfbi(10 dB)
log If
I0
log Ii
I0
bi(10 dB) log Ii
I0
bf(10 dB) log If
I0
Additional examples, video, and practice available at WileyPLUS
Tim Mosenfelder/Getty Images, Inc.
Figure 17-11 Lars Ulrich
of Metallica is an
advocate for the
organization HEAR
(Hearing Education
and Awareness for
Rockers), which warns
about the damage high
sound levels can have
on hearing.
17-5 SOURCES OF MUSICAL SOUND
Learning Objectives
17.26 Identify which type of pipe has even harmonics.
17.27 For any given harmonic and for a pipe with only one open
end or with two open ends, apply the relationships between
the pipe length L, the speed of sound v, the wavelength l, the
harmonic frequency f, and the harmonic number n.
After reading this module, you should be able to . . .
17.24 Using standing wave patterns for string waves, sketch the
standing wave patterns for the first several acoustical harmon-
ics of a pipe with only one open end and with two open ends.
17.25 For a standing wave of sound, relate the distance
between nodes and the wavelength.
●Standing sound wave patterns can be set up in pipes (that
is, resonance can be set up) if sound of the proper wave-
length is introduced in the pipe.
●A pipe open at both ends will resonate at frequencies
,n1,2,3,...,fv
nv
2L
where vis the speed of sound in the air in the pipe.
●For a pipe closed at one end and open at the other, the
resonant frequencies are
,n1,3,5,....fv
nv
4L
Key Ideas
493
17-5 SOURCES OF MUSICAL SOUND
(b)
λ
= 2L
L
A N A
(a)
Antinodes (maximum oscillation)
occur at the open ends.
First harmonic
Figure 17-13 (a) The simplest standing wave pattern of displacement for (longitudinal)
sound waves in a pipe with both ends open has an antinode (A) across each end and a
node (N) across the middle. (The longitudinal displacements represented by the double
arrows are greatly exaggerated.) (b) The corresponding standing wave pattern for
(transverse) string waves.
Sources of Musical Sound
Musical sounds can be set up by oscillating strings (guitar, piano, violin), mem-
branes (kettledrum, snare drum), air columns (flute, oboe, pipe organ, and the
didgeridoo of Fig. 17-12), wooden blocks or steel bars (marimba, xylophone), and
many other oscillating bodies. Most common instruments involve more than a
single oscillating part.
Recall from Chapter 16 that standing waves can be set up on a stretched
string that is fixed at both ends. They arise because waves traveling along the
string are reflected back onto the string at each end. If the wavelength of the
waves is suitably matched to the length of the string, the superposition of waves
traveling in opposite directions produces a standing wave pattern (or oscillation
mode). The wavelength required of the waves for such a match is one that cor-
responds to a resonant frequency of the string. The advantage of setting up
standing waves is that the string then oscillates with a large, sustained amplitude,
pushing back and forth against the surrounding air and thus generating a notice-
able sound wave with the same frequency as the oscillations of the string. This
production of sound is of obvious importance to, say, a guitarist.
Sound Waves. We can set up standing waves of sound in an air-filled pipe in
a similar way.As sound waves travel through the air in the pipe,they are reflected
at each end and travel back through the pipe. (The reflection occurs even if an
end is open, but the reflection is not as complete as when the end is closed.) If the
wavelength of the sound waves is suitably matched to the length of the pipe, the
superposition of waves traveling in opposite directions through the pipe sets up
a standing wave pattern. The wavelength required of the sound waves for such a
match is one that corresponds to a resonant frequency of the pipe. The advan-
tage of such a standing wave is that the air in the pipe oscillates with a large,
sustained amplitude, emitting at any open end a sound wave that has the same
frequency as the oscillations in the pipe. This emission of sound is of obvious
importance to, say, an organist.
Many other aspects of standing sound wave patterns are similar to those
of string waves: The closed end of a pipe is like the fixed end of a string in that
there must be a node (zero displacement) there, and the open end of a pipe is
like the end of a string attached to a freely moving ring, as in Fig. 16-19b,in
that there must be an antinode there. (Actually, the antinode for the open
end of a pipe is located slightly beyond the end, but we shall not dwell on
that detail.)
Two Open Ends. The simplest standing wave pattern that can be set up in a
pipe with two open ends is shown in Fig. 17-13a.There is an antinode across each
Alamy
Figure 17-12 The air column within a didgeri-
doo (“a pipe”) oscillates when the instru-
ment is played.
494 CHAPTER 17 WAVES—II
open end, as required. There is also a node across the middle of the pipe. An
easier way of representing this standing longitudinal sound wave is shown in
Fig. 17-13b—by drawing it as a standing transverse string wave.
The standing wave pattern of Fig. 17-13ais called the fundamental mode or
first harmonic. For it to be set up, the sound waves in a pipe of length Lmust
have a wavelength given by Ll/2, so that l2L. Several more standing
sound wave patterns for a pipe with two open ends are shown in Fig. 17-14a
using string wave representations. The second harmonic requires sound waves of
wavelength lL, the third harmonic requires wavelength l2L/3, and so on.
More generally, the resonant frequencies for a pipe of length Lwith two
open ends correspond to the wavelengths
for n1,2,3,..., (17-38)
where nis called the harmonic number. Letting vbe the speed of sound, we write
the resonant frequencies for a pipe with two open ends as
for n1,2,3,... (pipe,two open ends). (17-39)
One Open End. Figure 17-14bshows (using string wave representations)
some of the standing sound wave patterns that can be set up in a pipe with only
one open end.As required, across the open end there is an antinode and across
the closed end there is a node. The simplest pattern requires sound waves hav-
ing a wavelength given by Ll/4, so that l4L.The next simplest pattern re-
quires a wavelength given by L3l/4, so that l4L/3, and so on.
More generally, the resonant frequencies for a pipe of length Lwith only
one open end correspond to the wavelengths
for n1,3,5,..., (17-40)
in which the harmonic number n must be an odd number. The resonant frequen-
cies are then given by
for n1,3,5,... (pipe,one open end). (17-41)
Note again that only odd harmonics can exist in a pipe with one open end. For
example, the second harmonic, with n2, cannot be set up in such a pipe.
Note also that for such a pipe the adjective in a phrase such as “the third har-
monic” still refers to the harmonic number n(and not to, say, the third possible
harmonic). Finally note that Eqs. 17-38 and 17-39 for two open ends contain the
fv
lnv
4L,
l4L
n,
fv
lnv
2L,
l2L
n,
L
n= 2
Second
n= 3
Third
Fourth
n= 4
λ
= 2L/2 = L
λ
= 2L/3
λ
= 2L/4 = L/2
(a)Two open ends—
any harmonic
(b)
λ
= 4L
λ
= 4L/3
λ
= 4L/5
λ
= 4L/7
n= 1
First
Third
Fifth
Seventh
n= 3
n= 5
n= 7
One open end—
only odd harmonics
Figure 17-14 Standing wave patterns for string waves superimposed on pipes to represent
standing sound wave patterns in the pipes. (a) With both ends of the pipe open, any harmonic
can be set up in the pipe. (b) With only one end open, only odd harmonics can be set up.
495
17-5 SOURCES OF MUSICAL SOUND
number 2 and any integer value of n, but Eqs. 17-40 and 17-41 for one open end
contain the number 4 and only odd values of n.
Length. The length of a musical instrument reflects the range of frequencies
over which the instrument is designed to function, and smaller length implies
higher frequencies, as we can tell from Eq. 16-66 for string instruments and Eqs. 17-
39 and 17-41 for instruments with air columns. Figure 17-15, for example, shows the
saxophone and violin families, with their frequency ranges suggested by the piano
keyboard. Note that, for every instrument, there is overlap with its higher- and
lower-frequency neighbors.
Net Wave. In any oscillating system that gives rise to a musical sound,
whether it is a violin string or the air in an organ pipe, the fundamental and one
or more of the higher harmonics are usually generated simultaneously. Thus, you
hear them together—that is, superimposed as a net wave.When different instru-
ments are played at the same note, they produce the same fundamental fre-
quency but different intensities for the higher harmonics. For example, the fourth
harmonic of middle C might be relatively loud on one instrument and relatively
quiet or even missing on another. Thus, because different instruments produce
different net waves, they sound different to you even when they are played at the
same note. That would be the case for the two net waves shown in Fig. 17-16,
which were produced at the same note by different instruments. If you heard only
the fundamentals, the music would not be musical.
Figure 17-15 The saxophone and
violin families,showing the re-
lations between instrument
length and frequency range.
The frequency range of each
instrument is indicated by a
horizontal bar along a fre-
quency scale suggested by the
keyboard at the bottom; the
frequency increases toward
the right.
Figure 17-16 The wave forms produced by (a)
a flute and (b) an oboe when played at the
same note, with the same first harmonic
frequency.
Tim
e
(a)
(b)
ABCDEFGABCDEFGABCDEFGABCDEFGABCDEFGABCDEFGABCBCDEFGA
Bass saxophone
Soprano saxophone
Bass
Cello
Viola
Violin
Baritone saxophone
Tenor saxophone
Alto saxophone
Checkpoint 3
Pipe A, with length L, and pipe B, with length 2L, both have two open ends. Which
harmonic of pipe Bhas the same frequency as the fundamental of pipe A?
KEY IDEAS
(1) The sound from one pipe can set up a standing wave in
another pipe only if the harmonic frequencies match. (2)
Equation 17-39 gives the harmonic frequencies in a pipe
with two open ends (a symmetric pipe) as fnv/2L, for
n1,2,3,...,that is, for any positive integer. (3) Equation
Sample Problem 17.05 Resonance between pipes of different lengths
Pipe Ais open at both ends and has length LA0.343 m.
We want to place it near three other pipes in which standing
waves have been set up, so that the sound can set up a stand-
ing wave in pipe A. Those other three pipes are each closed
at one end and have lengths LB0.500LA,LC0.250LA,
and LD2.00LA. For each of these three pipes, which of
their harmonics can excite a harmonic in pipe A?
496 CHAPTER 17 WAVES—II
17-6 BEATS
After reading this module, you should be able to . . .
17.28 Explain how beats are produced.
17.29 Add the displacement equations for two sound
waves of the same amplitude and slightly different angu-
lar frequencies to find the displacement equation of the
resultant wave and identify the time-varying amplitude.
17.30 Apply the relationship between the beat frequency
and the frequencies of two sound waves that have the
same amplitude when the frequencies (or, equivalently,
the angular frequencies) differ by a small amount.
Learning Objectives
●Beats arise when two waves having slightly different frequencies, f1and f2, are detected together. The beat frequency is
fbeat f1f2.
Key Idea
Additional examples, video, and practice available at WileyPLUS
1
13
35
135
132465
0.500 1.51.0 2.0 3.02.5
7 9 11 13 15 17 19 21 23
nA
kHz
nB
nC
nD
Figure 17-17 Harmonic frequencies of four pipes.
17-41 gives the harmonic frequencies in a pipe with only
one open end (an asymmetric pipe) as fnv/4L, for n
1,3,5,...,that is, for only odd positive integers.
Pipe A: Let’s first find the resonant frequencies of symmet-
ric pipe A(with two open ends) by evaluating Eq. 17-39:
nA(500 Hz) nA(0.50 kHz), for nA1,2,3,....
The first six harmonic frequencies are shown in the top plot
in Fig.17-17.
Pipe B: Next let’s find the resonant frequencies of asym-
metric pipe B(with only one open end) by evaluating Eq.
17-41, being careful to use only odd integers for the har-
monic numbers:
nB(500 Hz) nB(0.500 kHz), for nB1,3,5,....
Comparing our two results, we see that we get a match for
each choice of nB:
fA=fBfor nAnBwith nB1,3,5,.... (Answer)
For example, as shown in Fig. 17-17, if we set up the fifth
harmonic in pipe Band bring the pipe close to pipe A, the
fifth harmonic will then be set up in pipe A. However, no
harmonic in Bcan set up an even harmonic in A.
Pipe C: Let’s continue with pipe C(with only one end) by
writing Eq. 17-41 as
fBnBv
4LB
nBv
4(0.500LA)nB(343 m/s)
2(0.343 m)
fAnAv
2LA
nA(343 m/s)
2(0.343 m)
nC(1000 Hz) nC(1.00 kHz), for nC1,3,5,....
From this we see that Ccan excite some of the harmonics of
Abut only those with harmonic numbers nAthat are twice
an odd integer:
fAfCfor nA2nC, with nC1,3,5,.... (Answer)
Pipe D: Finally, let’s check Dwith our same procedure:
nD(125 Hz) nD(0.125 kHz), for nD1,3,5,....
As shown in Fig. 17-17, none of these frequencies match a
harmonic frequency of A. (Can you see that we would get a
match if nD4nA? But that is impossible because 4nAcan-
not yield an odd integer, as required of nD.) Thus Dcannot
set up a standing wave in A.
fDnDv
4LD
nDv
4(2LA)nD(343 m/s)
8(0.343 m/s)
fCnCv
4LC
nCv
4(0.250LA)nC(343 m/s)
0.343 m/s
497
17-6 BEATS
Beats
If we listen, a few minutes apart, to two sounds whose frequencies are, say, 552
and 564 Hz, most of us cannot tell one from the other because the frequencies are
so close to each other. However, if the sounds reach our ears simultaneously,
what we hear is a sound whose frequency is 558 Hz, the average of the two com-
bining frequencies. We also hear a striking variation in the intensity of this
sound—it increases and decreases in slow, wavering beats that repeat at a fre-
quency of 12 Hz, the difference between the two combining frequencies. Figure
17-18 shows this beat phenomenon.
Let the time-dependent variations of the displacements due to two sound
waves of equal amplitude smbe
s1smcos v1tand s2smcos v2t, (17-42)
where v1v2. From the superposition principle, the resultant displacement is the
sum of the individual displacements:
ss1s2sm(cos v1tcos v2t).
Using the trigonometric identity (see Appendix E)
allows us to write the resultant displacement as
. (17-43)
If we write
v (v1v2) and v(v1v2), (17-44)
we can then write Eq. 17-43 as
s(t)[2smcos vt] cos vt. (17-45)
We now assume that the angular frequencies v1and v2of the combining
waves are almost equal, which means that vvin Eq. 17-44. We can then
regard Eq. 17-45 as a cosine function whose angular frequency is vand whose
amplitude (which is not constant but varies with angular frequency v) is the
absolute value of the quantity in the brackets.
A maximum amplitude will occur whenever cos vtin Eq. 17-45 has the
value 1 or 1, which happens twice in each repetition of the cosine function.
Because cos vthas angular frequency v, the angular frequency vbeat at which
beats occur is vbeat 2v. Then, with the aid of Eq. 17-44, we can write the beat
angular frequency as
.
Because v2pf, we can recast this as
fbeat f1f2(beat frequency). (17-46)
Musicians use the beat phenomenon in tuning instruments. If an instrument is
sounded against a standard frequency (for example, the note called “concert A”
played on an orchestra’s first oboe) and tuned until the beat disappears, the instru-
ment is in tune with that standard. In musical Vienna, concert A (440 Hz) is avail-
able as a convenient telephone service for the city’s many musicians.
vbeat 2v(2)(1
2)(v1v2)v1v2
1
2
1
2
s2smcos[1
2(v1v2)t] cos[1
2(v1v2)t]
cos acos b2 cos[1
2(ab)] cos[1
2(ab)]
Figure 17-18 (a,b) The pressure variations pof two sound waves as they would be detected
separately.The frequencies of the waves are nearly equal. (c) The resultant pressure varia-
tion if the two waves are detected simultaneously.
Time
(c)
(b)
(a)
498 CHAPTER 17 WAVES—II
sociated with the second harmonics is
fbeat,2 fA2fB22fA12fB1
2(432 Hz) 2(371 Hz)
122 Hz. (Answer)
Experiments indicate that penguins can perceive such large
beat frequencies. (Humans cannot hear a beat frequency
any higher than about 12 Hz — we perceive the two sepa-
rate frequencies.) Thus, a penguin’s cry can be rich with dif-
ferent harmonics and different beat frequencies, allowing
the voice to be recognized even among the voices of thou-
sands of other,closely huddled penguins.
Additional examples, video, and practice available at WileyPLUS
17-7 THE DOPPLER EFFECT
After reading this module, you should be able to . . .
17.31 Identify that the Doppler effect is the shift in the de-
tected frequency from the frequency emitted by a sound
source due to the relative motion between the source and
the detector.
17.32 Identify that in calculating the Doppler shift in sound,
the speeds are measured relative to the medium (such as
air or water), which may be moving.
17.33 Calculate the shift in sound frequency for (a) a source
moving either directly toward or away from a stationary
detector, (b) a detector moving either directly toward or
away from a stationary source, and (c) both source and
detector moving either directly toward each other or
directly away from each other.
17.34 Identify that for relative motion between a sound source
and a sound detector, motion toward tends to shift the
frequency up and motion away tends to shift it down.
Learning Objectives
●The Doppler effect is a change in the observed frequency of
a wave when the source or the detector moves relative to the
transmitting medium (such as air). For sound the observed fre-
quency fis given in terms of the source frequency fby
(general Doppler effect),ffvvD
vvS
Key Ideas
Because the standing waves in the penguin are effec-
tively in a pipe with two open ends, the resonant frequencies
are given by Eq. 17-39 (fnv/2L), in which Lis the
(unknown) length of the effective pipe. The first-harmonic
frequency is f1v/2L, and the second-harmonic frequency
is f22v/2L. Comparing these two frequencies, we see that,
in general,
f22f1.
For the penguin, the second harmonic of side Ahas
frequency fA22fA1and the second harmonic of side Bhas
frequency fB22fB1. Using Eq. 17-46 with frequencies
fA2and fB2, we find that the corresponding beat frequency as-
Sample Problem 17.06 Beat frequencies and penguins finding one another
When an emperor penguin returns from a search for food,
how can it find its mate among the thousands of penguins
huddled together for warmth in the harsh Antarctic
weather? It is not by sight, because penguins all look alike,
even to a penguin.
The answer lies in the way penguins vocalize. Most birds
vocalize by using only one side of their two-sided vocal or-
gan, called the syrinx. Emperor penguins, however, vocalize
by using both sides simultaneously. Each side sets up
acoustic standing waves in the bird’s throat and mouth,
much like in a pipe with two open ends. Suppose that the
frequency of the first harmonic produced by side Ais fA1
432 Hz and the frequency of the first harmonic produced by
side Bis fB1371 Hz. What is the beat frequency between
those two first-harmonic frequencies and between the two
second-harmonic frequencies?
KEY IDEA
The beat frequency between two frequencies is their differ-
ence,as given by Eq.17-46 ( fbeat f1f2).
Calculations: For the two first-harmonic frequencies fA1
and fB1, the beat frequency is
fbeat,1 fA1fB1432 Hz 371 Hz
61 Hz. (Answer)
where vDis the speed of the detector relative to the medium,
vSis that of the source, and vis the speed of sound in the
medium.
●The signs are chosen such that ftends to be greater for
relative motion toward (one of the objects moves toward the
other) and less for motion away.
499
17-7 THE DOPPLER EFFECT
The Doppler Effect
A police car is parked by the side of the highway, sounding its 1000 Hz siren. If
you are also parked by the highway, you will hear that same frequency.
However, if there is relative motion between you and the police car, either
toward or away from each other, you will hear a different frequency. For exam-
ple, if you are driving toward the police car at 120 km/h (about 75 mi/h), you will
hear a higher frequency (1096 Hz, an increase of 96 Hz). If you are driving away
from the police car at that same speed, you will hear a lower frequency (904 Hz,
adecrease of 96 Hz).
These motion-related frequency changes are examples of the Doppler
effect. The effect was proposed (although not fully worked out) in 1842 by
Austrian physicist Johann Christian Doppler. It was tested experimentally in
1845 by Buys Ballot in Holland, “using a locomotive drawing an open car with
several trumpeters.”
The Doppler effect holds not only for sound waves but also for electromag-
netic waves, including microwaves, radio waves, and visible light. Here, however,
we shall consider only sound waves, and we shall take as a reference frame the
body of air through which these waves travel. This means that we shall measure
the speeds of a source Sof sound waves and a detector Dof those waves relative
to that body of air. (Unless otherwise stated, the body of air is stationary relative
to the ground, so the speeds can also be measured relative to the ground.)
We shall assume that Sand Dmove either directly toward or directly away from
each other,at speeds less than the speed of sound.
General Equation. If either the detector or the source is moving, or both are
moving,the emitted frequency fand the detected frequency fare related by
(general Doppler effect), (17-47)
where vis the speed of sound through the air, vDis the detector’s speed relative
to the air, and vSis the source’s speed relative to the air. The choice of plus or
minus signs is set by this rule:
ffvvD
vvS
When the motion of detector or source is toward the other, the sign on its speed
must give an upward shift in frequency. When the motion of detector or source is
away from the other, the sign on its speed must give a downward shift in frequency.
In short, toward means shift up, and away means shift down.
Here are some examples of the rule. If the detector moves toward the
source, use the plus sign in the numerator of Eq. 17-47 to get a shift up in the
frequency. If it moves away, use the minus sign in the numerator to get a shift
down. If it is stationary, substitute 0 for vD. If the source moves toward the
detector, use the minus sign in the denominator of Eq. 17-47 to get a shift up in
the frequency. If it moves away, use the plus sign in the denominator to get
a shift down. If the source is stationary, substitute 0 for vS.
Next, we derive equations for the Doppler effect for the following two
specific situations and then derive Eq. 17-47 for the general situation.
1. When the detector moves relative to the air and the source is stationary
relative to the air, the motion changes the frequency at which the detector
intercepts wavefronts and thus changes the detected frequency of the sound
wave.
2. When the source moves relative to the air and the detector is stationary
relative to the air, the motion changes the wavelength of the sound wave and
thus changes the detected frequency (recall that frequency is related to
wavelength).
500 CHAPTER 17 WAVES—II
Detector Moving, Source Stationary
In Fig. 17-19, a detector D(represented by an ear) is moving at speed vDtoward
a stationary source Sthat emits spherical wavefronts, of wavelength land
frequency f, moving at the speed vof sound in air.The wavefronts are drawn one
wavelength apart. The frequency detected by detector Dis the rate at which D
intercepts wavefronts (or individual wavelengths). If Dwere stationary, that rate
would be f, but since Dis moving into the wavefronts, the rate of interception is
greater,and thus the detected frequency fis greater than f.
Let us for the moment consider the situation in which Dis stationary
(Fig. 17-20). In time t, the wavefronts move to the right a distance vt. The num-
ber of wavelengths in that distance vt is the number of wavelengths intercepted
by Din time t, and that number is vt/l. The rate at which Dintercepts wave-
lengths, which is the frequency fdetected by D,is
(17-48)
In this situation, with Dstationary, there is no Doppler effect—the frequency
detected by Dis the frequency emitted by S.
Now let us again consider the situation in which Dmoves in the direction
opposite the wavefront velocity (Fig. 17-21). In time t, the wavefronts move to
the right a distance vt as previously, but now Dmoves to the left a distance vDt.
Thus, in this time t, the distance moved by the wavefronts relative to Dis vt
vDt.The number of wavelengths in this relative distance vt vDtis the number
of wavelengths intercepted by Din time tand is (vt vDt)/l. The rate at which
Dintercepts wavelengths in this situation is the frequency f,given by
. (17-49)
From Eq. 17-48, we have lv/f.Then Eq. 17-49 becomes
. (17-50)
Note that in Eq. 17-50, f>funless vD0 (the detector is stationary).
Similarly, we can find the frequency detected by Dif Dmoves away from
the source. In this situation, the wavefronts move a distance vt vDtrelative to
Din time t, and fis given by
. (17-51)
In Eq. 17-51, f<funless vD0.We can summarize Eqs. 17-50 and 17-51 with
(detector moving,source stationary). (17-52)f fvvD
v
f fvvD
v
f vvD
v/ffvvD
v
f (vt vDt)/
tvvD
fvt/l
tv
l.
Figure 17-19 A stationary source of
sound Semits spherical wavefronts,
shown one wavelength apart, that
expand outward at speed v. A sound
detector D, represented by an ear,
moves with velocity toward the
source. The detector senses a higher
frequency because of its motion.
v
:
D
λ
vS= 0
Sx
D
λ
vD
vv
Shift up: The detector
moves toward the source.
Figure 17-20 The wavefronts of Fig. 17-19,
assumed planar, (a) reach and (b) pass
a stationary detector D; they move a
distance vt to the right in time t.
v
v
D
vt
(a)
(b)
λ
Figure 17-21 Wavefronts traveling to the
right (a) reach and (b) pass detector D,
which moves in the opposite direction. In
time t, the wavefronts move a distance vt
to the right and Dmoves a distance vDtto
the left.
v
v
vt
(a)
(b)
λ
D
vDt
vD
vD
501
17-7 THE DOPPLER EFFECT
Source Moving, Detector Stationary
Let detector Dbe stationary with respect to the body of air, and let source S
move toward Dat speed vS(Fig. 17-22).The motion of Schanges the wavelength
of the sound waves it emits and thus the frequency detected by D.
To see this change, let T( 1/f) be the time between the emission of any
pair of successive wavefronts W1and W2. During T, wavefront W1moves a dis-
tance vT and the source moves a distance vST. At the end of T, wavefront W2is
emitted. In the direction in which Smoves, the distance between W1and W2, which
is the wavelength lof the waves moving in that direction, is vT vST. If Ddetects
those waves, it detects frequency fgiven by
. (17-53)
Note that fmust be greater than funless vS0.
In the direction opposite that taken by S, the wavelength lof the waves is
again the distance between successive waves but now that distance is vT vST.If
Ddetects those waves, it detects frequency fgiven by
. (17-54)
Now fmust be less than funless vS0.
We can summarize Eqs. 17-53 and 17-54 with
(source moving,detector stationary). (17-55)
General Doppler Effect Equation
We can now derive the general Doppler effect equation by replacing fin
Eq. 17-55 (the source frequency) with fof Eq. 17-52 (the frequency associated
with motion of the detector). That simple replacement gives us Eq. 17-47 for the
general Doppler effect.That general equation holds not only when both detector
and source are moving but also in the two specific situations we just discussed.
For the situation in which the detector is moving and the source is stationary, sub-
stitution of vS0 into Eq. 17-47 gives us Eq. 17-52, which we previously found.
For the situation in which the source is moving and the detector is stationary,
substitution of vD0 into Eq. 17-47 gives us Eq. 17-55, which we previously
found.Thus, Eq. 17-47 is the equation to remember.
ffv
vvS
ffv
vvS
fv
vvS
f v
v
vT vSTv
v/fvS/f
Figure 17-22 A detector Dis stationary,
and a source Sis moving toward it at
speed vS. Wavefront W1was emitted
when the source was at S1, wavefront
W7when it was at S7. At the moment
depicted, the source is at S.The
detector senses a higher frequency
because the moving source, chasing
its own wavefronts, emits a reduced
wavelength lin the direction of its
motion.
W1
x
λ
'
SS7
S1
vS
W7
W2
vD = 0
D
Shift up: The source moves
toward the detector.
502 CHAPTER 17 WAVES—II
numerator, we choose the minus sign to meet that tendency
(the numerator becomes smaller). These reasoning steps
are shown in Table 17-3.
We have the speed of the bat in the denominator of
Eq. 17-56. The bat moves toward the moth, which tends to
increase the detected frequency. Because the speed is in the
denominator, we choose the minus sign to meet that ten-
dency (the denominator becomes smaller).
With these substitutions and decisions, we have
(Answer)
Detection of echo by bat: In the echo back to the bat, the
moth acts as a source of sound, emitting at the frequency fmd
we just calculated. So now the moth is the source (moving
away) and the bat is the detector (moving toward).The rea-
soning steps are shown in Table 17-3. To find the frequency
fbd detected by the bat, we write Eq. 17-56 as
(Answer)
Some moths evade bats by “jamming” the detection system
with ultrasonic clicks.
83.00 kHz 83.0 kHz.
(82.767 kHz) 343 m/s 9.00 m/s
343 m/s 8.00 m/s
fbd fmd
vvb
vvm
82.767 kHz 82.8 kHz.
(82.52 kHz) 343 m/s 8.00 m/s
343 m/s 9.00 m/s
fmd fbe
vvm
vvb
Sample Problem 17.07 Double Doppler shift in the echoes used by bats
Bats navigate and search out prey by emitting, and then de-
tecting reflections of, ultrasonic waves, which are sound
waves with frequencies greater than can be heard by a hu-
man. Suppose a bat emits ultrasound at frequency fbe 82.52
kHz while flying with velocity as it chases a
moth that flies with velocity . What fre-
quency fmd does the moth detect? What frequency fbd does the
bat detect in the returning echo from the moth?
KEY IDEAS
The frequency is shifted by the relative motion of the bat and
moth. Because they move along a single axis, the shifted fre-
quency is given by Eq. 17-47. Motion toward tends to shift the
frequency up, and motion away tends to shift it down.
Detection by moth: The general Doppler equation is
(17-56)
Here, the detected frequency fthat we want to find is the fre-
quency fmd detected by the moth. On the right side, the emitted
frequency fis the bat’s emission frequency fbe 82.52 kHz, the
speed of sound is v343 m/s, the speed vDof the detector is
the moth’s speed vm8.00 m/s, and the speed vSof the
source is the bat’s speed vb9.00 m/s.
The decisions about the plus and minus signs can be
tricky. Think in terms of toward and away. We have the
speed of the moth (the detector) in the numerator of
Eq. 17-56. The moth moves away from the bat, which tends
to lower the detected frequency. Because the speed is in the
ffvvD
vvS
.
v
:
m(8.00 m/s)i
ˆ
v
:
b(9.00 m/s)i
ˆ
Additional examples, video, and practice available at WileyPLUS
Table 17-3
Bat to Moth
Detector Source
moth bat
speed vDvmspeed vSvb
away toward
shift down shift up
numerator denominator
minus minus
Echo Back to Bat
Detector Source
bat moth
speed vDvbspeed vSvm
toward away
shift up shift down
numerator denominator
plus plus
Checkpoint 4
The figure indicates the directions of motion of a sound source and a detector for
six situations in stationary air.For each situation, is the detected frequency greater
than or less than the emitted frequency, or can’t we tell without more information
about the actual speeds?
Source Detector Source Detector
(a)99: • 0 speed(d);99 ;99
(b);99 • 0 speed (e)99: ;99
(c)99: 99: (f);99 99:
503
17-8 SUPERSONIC SPEEDS, SHOCK WAVES
Supersonic Speeds, Shock Waves
If a source is moving toward a stationary detector at a speed vSequal to
the speed of soundv, Eqs. 17-47 and 17-55 predict that the detected fre-
quency fwill be infinitely great. This means that the source is moving so
fast that it keeps pace with its own spherical wavefronts (Fig. 17-23a).
What happens when vS> v? For such supersonic speeds, Eqs. 17-47 and
17-55 no longer apply. Figure 17-23bdepicts the spherical wavefronts that
originated at various positions of the source. The radius of any wavefront
is vt, where tis the time that has elapsed since the source emitted that
wavefront. Note that all the wavefronts bunch along a V-shaped envelope
in this two-dimensional drawing. The wavefronts actually extend in
three dimensions, and the bunching actually forms a cone called the Mach
cone.Ashock wave exists along the surface of this cone, because the
bunching of wavefronts causes an abrupt rise and fall of air pressure as
the surface passes through any point. From Fig. 17-23b, we see that the
half-angle uof the cone (the Mach cone angle) is given by
(Mach cone angle). (17-57)
The ratio vS/vis the Mach number. If a plane flies at Mach 2.3, its
speed is 2.3 times the speed of sound in the air through which the plane
is flying. The shock wave generated by a supersonic aircraft (Fig. 17-24)
sin
vt
vStv
vS
17-8 SUPERSONIC SPEEDS, SHOCK WAVES
After reading this module, you should be able to . . .
17.35 Sketch the bunching of wavefronts for a sound source
traveling at the speed of sound or faster.
17.36 Calculate the Mach number for a sound source
exceeding the speed of sound.
17.37 For a sound source exceeding the speed of
sound, apply the relationship between the Mach
cone angle, the speed of sound, and the speed of the
source.
Learning Objectives
●If the speed of a source relative to the medium exceeds the speed of sound in the medium, the Doppler equation no longer
applies. In such a case, shock waves result. The half-angle uof the Mach cone is given by
(Mach cone angle).sin
v
vS
Key Idea
Figure 17-24 Shock waves produced by the wings of a Navy FA
18 jet.The shock waves are visible because the sudden decrease
in air pressure in them caused water molecules in the air to con-
dense, forming a fog.
U.S. Navy photo by Ensign John Gay
Figure 17-23 (a) A source of sound Smoves at speed
vSequal to the speed of sound and thus as fast as the
wavefronts it generates. (b) A source Smoves at
speed vSfaster than the speed of sound and thus
faster than the wavefronts.When the source was at
position S1it generated wavefront W1, and at posi-
tion S6it generated W6. All the spherical wavefronts
expand at the speed of sound vand bunch along the
surface of a cone called the Mach cone, forming a
shock wave. The surface of the cone has half-angle u
and is tangent to all the wavefronts.
Surface of
Mach cone
W6
W1
x
(b)
S
S6
S1
θ
vS
vSt
vt
x
vS
(a)
S
504 CHAPTER 17 WAVES—II
or projectile produces a burst of sound, called a sonic boom, in which the air pres-
sure first suddenly increases and then suddenly decreases below normal before re-
turning to normal. Part of the sound that is heard when a rifle is fired is the sonic
boom produced by the bullet.When a long bull whip is snapped, its tip is moving
faster than sound and produces a small sonic boom—the crack of the whip.
Sound Waves Sound waves are longitudinal mechanical waves
that can travel through solids, liquids, or gases. The speed vof a
sound wave in a medium having bulk modulus Band density ris
(speed of sound). (17-3)
In air at 20C,the speed of sound is 343 m/s.
A sound wave causes a longitudinal displacement sof a mass
element in a medium as given by
ssmcos(kx vt), (17-12)
where smis the displacement amplitude (maximum displacement)
from equilibrium, k2p/l, and v2pf,land fbeing the wave-
length and frequency of the sound wave. The wave also causes a
pressure change pfrom the equilibrium pressure:
ppmsin(kx vt), (17-13)
where the pressure amplitude is
pm(vrv)sm. (17-14)
Interference The interference of two sound waves with identi-
cal wavelengths passing through a common point depends on their
phase difference fthere. If the sound waves were emitted in phase
and are traveling in approximately the same direction, fis given by
(17-21)
where Lis their path length difference (the difference in the
distances traveled by the waves to reach the common point). Fully
constructive interference occurs when fis an integer multiple of 2p,
fm(2p), for m0,1,2,..., (17-22)
and, equivalently, when Lis related to wavelength lby
0,1,2,.... (17-23)
Fully destructive interference occurs when fis an odd multiple of p,
f(2m1)p, for m0,1,2,..., (17-24)
and, equivalently, when Lis related to lby
0.5, 1.5, 2.5,.... (17-25)
Sound Intensity The intensity Iof a sound wave at a surface is
the average rate per unit area at which energy is transferred by the
wave through or onto the surface:
, (17-26)
where Pis the time rate of energy transfer (power) of the sound wave
IP
A
L
l
L
l
fL
l 2p,
vAB
Review & Summary
and Ais the area of the surface intercepting the sound.The intensity I
is related to the displacement amplitude smof the sound wave by
(17-27)
The intensity at a distance rfrom a point source that emits sound
waves of power Psis
. (17-28)
Sound Level in Decibels The sound level bin decibels (dB)
is defined as
, (17-29)
where I0(1012 W/m2) is a reference intensity level to which all
intensities are compared. For every factor-of-10 increase in inten-
sity, 10 dB is added to the sound level.
Standing Wave Patterns in Pipes Standing sound wave
patterns can be set up in pipes. A pipe open at both ends will
resonate at frequencies
,n1,2,3,..., (17-39)
where vis the speed of sound in the air in the pipe. For a pipe
closed at one end and open at the other, the resonant fre-
quencies are
,n1,3,5,.... (17-41)
Beats Beats arise when two waves having slightly different fre-
quencies, f1and f2, are detected together.The beat frequency is
fbeat f1f2. (17-46)
The Doppler Effect The Doppler effect is a change in the
observed frequency of a wave when the source or the detec-
tor moves relative to the transmitting medium (such as air).
For sound the observed frequency fis given in terms of the source
frequency fby
(general Doppler effect), (17-47)
where vDis the speed of the detector relative to the medium, vSis
that of the source, and vis the speed of sound in the medium. The
signs are chosen such that ftends to be greater for motion toward
and less for motion away.
Shock Wave If the speed of a source relative to the medium
exceeds the speed of sound in the medium, the Doppler equation
no longer applies. In such a case, shock waves result.The half-angle
uof the Mach cone is given by
(Mach cone angle). (17-57)sin uv
vS
ffvvD
vvS
fv
nv
4L
fv
nv
2L
b(10 dB) log I
I0
IPs
4pr2
I1
2
vv2s2
m.
505
QUESTIONS
6Pipe Ahas length Land one open end. Pipe Bhas length 2L
and two open ends. Which harmonics of pipe Bhave a frequency
that matches a resonant frequency of pipe A?
7Figure 17-28 shows a moving sound source Sthat emits at a certain
frequency, and four stationary sound detectors. Rank the detectors
according to the frequency of the sound they detect from the
source, greatest first.
1In a first experiment, a sinusoidal sound wave is sent through
a long tube of air, transporting energy at the average rate of Pavg,1.
In a second experiment, two other sound waves, identical to the
first one, are to be sent simultaneously through the tube with a
phase difference fof either 0, 0.2 wavelength, or 0.5 wavelength
between the waves. (a) With only mental calculation, rank those
choices of faccording to the average rate at which the waves will
transport energy, greatest first. (b) For the first choice of f, what is
the average rate in terms of Pavg,1?
2In Fig. 17-25, two point sources
S1and S2, which are in phase, emit
identical sound waves of wave-
length 2.0 m. In terms of wave-
lengths, what is the phase differ-
ence between the waves arriving at
point Pif (a) L138 m and L234 m, and (b) L139 m and
L236 m? (c) Assuming that the source separation is much
smaller than L1and L2, what type of interference occurs at Pin
situations (a) and (b)?
3In Fig. 17-26, three long tubes
(A,B, and C) are filled with differ-
ent gases under different pressures.
The ratio of the bulk modulus to the
density is indicated for each gas in
terms of a basic value B0/r0. Each
tube has a piston at its left end
that can send a sound pulse
through the tube (as in Fig. 16-2).
The three pulses are sent simulta-
neously. Rank the tubes according
to the time of arrival of the pulses
at the open right ends of the tubes,
earliest first.
4The sixth harmonic is set up in a
pipe. (a) How many open ends does the pipe have (it has at least
one)? (b) Is there a node, antinode, or some intermediate state at the
midpoint?
5In Fig. 17-27, pipe Ais made to oscillate in its third harmonic
by a small internal sound source. Sound emitted at the right end
happens to resonate four nearby pipes, each with only one open
end (they are not drawn to scale). Pipe Boscillates in its lowest
harmonic, pipe Cin its second lowest harmonic, pipe Din its
third lowest harmonic, and pipe Ein its fourth lowest harmonic.
Without computation, rank all five pipes according to their
length, greatest first. (Hint: Draw the standing waves to scale and
then draw the pipes to scale.)
Questions
Figure 17-25 Question 2.
S1
S2
L1
L2
P
16B0/0
ρ
4B0/0
ρ
B0/0
ρ
A
B
C
L
L
L
LL
Figure 17-26 Question 3.
A
E
D
C
B
Figure 17-27 Question 5.
Figure 17-29 Question 8.
t
f
3
21
3
2 1
4
S
Figure 17-28 Question 7.
9For a particular tube, here are four of the six harmonic
frequencies below 1000 Hz: 300, 600, 750, and 900 Hz. What two
frequencies are missing from the list?
10 Figure 17-30 shows a stretched string of length Land pipes a,
b,c, and dof lengths L,2L,L/2, and L/2, respectively. The string’s
tension is adjusted until the speed of waves on the string equals the
speed of sound waves in the air. The fundamental mode of oscilla-
tion is then set up on the string. In which pipe will the sound pro-
duced by the string cause resonance, and what oscillation mode will
that sound set up?
L
b
dca
Figure 17-30 Question 10.
8A friend rides, in turn, the rims of three fast merry-go-rounds
while holding a sound source that emits isotropically at a certain
frequency. You stand far from each merry-go-round. The frequency
you hear for each of your friend’s three rides varies as the merry-go-
round rotates. The variations in frequency for the three rides are
given by the three curves in Fig. 17-29. Rank the curves according to
(a) the linear speed vof the sound source, (b) the angular speeds vof
the merry-go-rounds, and (c) the radii rof the merry-go-rounds,
greatest first.
11 You are given four tuning forks. The fork with the lowest fre-
quency oscillates at 500 Hz. By striking two tuning forks at a time,
you can produce the following beat frequencies, 1, 2, 3, 5, 7, and 8
Hz. What are the possible frequencies of the other three forks?
(There are two sets of answers.)
506 CHAPTER 17 WAVES—II
Rather, they change the value of dV/dp —that is, the differential
change in volume due to the differential change in the pressure
caused by the sound wave in the water. If fs/fi0.333, what is the
ratio (dV/dp)s/(dV/dp)i?
Module 17-2 Traveling Sound Waves
•9 If the form of a sound wave traveling through air is
s(x,t)(6.0 nm) cos(kx (3000 rad/s)tf),
how much time does any given air molecule along the path take to
move between displacements s2.0 nm and s2.0 nm?
•10 Underwater illusion. One
clue used by your brain to determine
the direction of a source of sound is
the time delay tbetween the arrival
of the sound at the ear closer to the
source and the arrival at the farther
ear.Assume that the source is distant
so that a wavefront from it is approx-
imately planar when it reaches you,
and let Drepresent the separation
between your ears. (a) If the source is located at angle uin front of
you (Fig. 17-31), what is in terms of Dand the speed of sound v
in air? (b) If you are submerged in water and the sound source is di-
rectly to your right, what is tin terms of Dand the speed of sound
vwin water? (c) Based on the time-delay clue, your brain interprets
the submerged sound to arrive at an angle ufrom the forward direc-
tion. Evaluate ufor fresh water at 20C.
•11 Diagnostic ultrasound of frequency 4.50 MHz is used to
examine tumors in soft tissue. (a) What is the wavelength in air of
such a sound wave? (b) If the speed of sound in tissue is 1500 m/s,
what is the wavelength of this wave in tissue?
•12 The pressure in a traveling sound wave is given by the
equation
p(1.50 Pa) sin p[(0.900 m1)x(315 s1)t].
Find the (a) pressure amplitude, (b) frequency, (c) wavelength, and
(d) speed of the wave.
••13 A sound wave of the form ssmcos(kx vtf) travels at
343 m/s through air in a long horizontal tube. At one instant, air
molecule Aat x2.000 m is at
its maximum positive displace-
ment of 6.00 nm and air mole-
cule Bat x2.070 m is at a pos-
itive displacement of 2.00 nm.
All the molecules between A
and Bare at intermediate dis-
placements. What is the fre-
quency of the wave?
••14 Figure 17-32 shows the
output from a pressure monitor
mounted at a point along the
SSM
t
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
•–••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Where needed in the problems, use
speed of sound in air 343 m/s
and density of air 1.21 kg/m3
unless otherwise specified.
Module 17-1 Speed of Sound
•1 Two spectators at a soccer game see, and a moment later hear,
the ball being kicked on the playing field. The time delay for spec-
tator Ais 0.23 s, and for spectator Bit is 0.12 s. Sight lines from the
two spectators to the player kicking the ball meet at an angle of
90. How far are (a) spectator Aand (b) spectator Bfrom the
player? (c) How far are the spectators from each other?
•2 What is the bulk modulus of oxygen if 32.0 g of oxygen occupies
22.4 L and the speed of sound in the oxygen is 317 m/s?
•3 When the door of the Chapel of the Mausoleum in
Hamilton, Scotland, is slammed shut, the last echo heard by some-
one standing just inside the door reportedly comes 15 s later. (a) If
that echo were due to a single reflection off a wall opposite the
door, how far from the door is the wall? (b) If, instead, the wall is
25.7 m away, how many reflections (back and forth) occur?
•4 A column of soldiers, marching at 120 paces per minute, keep
in step with the beat of a drummer at the head of the column. The
soldiers in the rear end of the column are striding forward with the
left foot when the drummer is advancing with the right foot.What is
the approximate length of the column?
••5 Earthquakes generate sound waves inside Earth.
Unlike a gas, Earth can experience both transverse (S) and longitu-
dinal (P) sound waves. Typically, the speed of S waves is about
4.5 km/s, and that of P waves 8.0 km/s. A seismograph records
P and S waves from an earthquake.The first P waves arrive 3.0 min
before the first S waves. If the waves travel in a straight line, how
far away did the earthquake occur?
••6 A man strikes one end of a thin rod with a hammer.
The speed of sound in the rod is 15 times the speed of sound in air.
A woman, at the other end with her ear close to the rod, hears the
sound of the blow twice with a 0.12 s interval between; one sound
comes through the rod and the other comes through the air along-
side the rod. If the speed of sound in air is 343 m/s, what is the
length of the rod?
••7 A stone is dropped into a well. The splash is
heard 3.00 s later.What is the depth of the well?
••8 Hot chocolate effect. Tap a metal spoon inside a
mug of water and note the frequency fiyou hear. Then add a
spoonful of powder (say, chocolate mix or instant coffee) and tap
again as you stir the powder. The frequency you hear has a lower
value fsbecause the tiny air bubbles released by the powder
change the water’s bulk modulus. As the bubbles reach the water
surface and disappear, the frequency gradually shifts back to its
initial value. During the effect, the bubbles don’t appreciably
change the water’s density or volume or the sound’s wavelength.
WWWSSM
ILWSSM
Figure 17-31 Problem 10.
DR
L
θ
θ
d
Wavefronts
Figure 17-32 Problem 14.
–1 1 2
t(ms)
Δp (mPa)
Δps
–Δps
separated by distance d12.00 m are
in phase. Assume the amplitudes of
the sound waves from the speakers
are approximately the same at the lis-
tener’s ear at distance d23.75 m di-
rectly in front of one speaker.
Consider the full audible range for
normal hearing, 20 Hz to 20 kHz. (a)
What is the lowest frequency fmin,1
that gives minimum signal (destructive interference) at the lis-
tener’s ear? By what number must fmin,1 be multiplied to get (b)
the second lowest frequency fmin,2 that gives minimum signal and
(c) the third lowest frequency fmin,3 that gives minimum signal?
(d) What is the lowest frequency fmax,1 that gives maximum signal
(constructive interference) at the listener’s ear? By what number
must fmax,1 be multiplied to get (e) the second lowest frequency
fmax,2 that gives maximum signal and (f) the third lowest fre-
quency fmax,3 that gives maximum signal?
••22 In Fig. 17-38, sound with a
40.0 cm wavelength travels right-
ward from a source and through a
tube that consists of a straight por-
tion and a half-circle. Part of the
sound wave travels through the half-
circle and then rejoins the rest of the
wave, which goes directly through
the straight portion. This rejoining
results in interference. What is the
smallest radius rthat results in an in-
tensity minimum at the detector?
•••23 Figure 17-39 shows two
point sources S1and S2that emit
sound of wavelength l2.00 m.
The emissions are isotropic and in
phase, and the separation between
507
PROBLEMS
path taken by a sound wave of a single frequency traveling at 343
m/s through air with a uniform density of 1.21 kg/m3. The vertical
axis scale is set by ps4.0 mPa.If the displacement function of the
wave is s(x,t)smcos(kx vt), what are (a) sm, (b) k, and (c) v?
The air is then cooled so that its density is 1.35 kg/m3and the speed
of a sound wave through it is 320 m/s.The sound source again emits
the sound wave at the same frequency and same pressure ampli-
tude.What now are (d) sm, (e) k,and (f) v?
••15 A handclap on stage in an amphitheater sends out
ql.What are the (a) smallest and (b) second smallest values of qthat
put Aand Bexactly out of phase with each other after the
reflections?
••19 Figure 17-35 shows two
Figure 17-33 Problem 15.
w
Terrace
isotropic point sources of sound, S1
and S2. The sources emit waves in
phase at wavelength 0.50 m; they are
separated by D1.75 m. If we move a sound detector along a large
circle centered at the midpoint between the sources, at how many
points do waves arrive at the detector (a) exactly in phase and (b) ex-
actly out of phase?
••20 Figure 17-36 shows four isotropic point sources of sound
that are uniformly spaced on an xaxis. The sources emit sound at
the same wavelength land same amplitude sm, and they emit in
phase. A point Pis shown on the xaxis. Assume that as the sound
waves travel to P, the decrease in their amplitude is negligible.
What multiple of smis the amplitude of the net wave at Pif dis-
tance din the figure is (a) l/4, (b) l/2, and (c) l?
••21 In Fig. 17-37, two speakers
SSM
sound waves that scatter from terraces of width w0.75 m
(Fig. 17-33). The sound returns to the stage as a periodic
series of pulses, one from each terrace; the parade of pulses
sounds like a played note. (a) Assuming that all the rays in
Fig. 17-33 are horizontal, find the frequency at which the pulses
return (that is, the frequency of the perceived note). (b) If the
width wof the terraces were smaller, would the frequency be
higher or lower?
Module 17-3 Interference
•16 Two sound waves, from two different sources with the same
frequency, 540 Hz, travel in the same direction at 330 m/s. The
sources are in phase.What is the phase difference of the waves at
a point that is 4.40 m from one source and 4.00 m from the
other?
••17 Two loud speakers are located 3.35 m apart on an
outdoor stage. A listener is 18.3 m from one and 19.5 m from the
other. During the sound check, a signal generator drives the two
speakers in phase with the same amplitude and frequency.
The transmitted frequency is swept through the audible range
(20 Hz to 20 kHz). (a) What is the lowest frequency fmin,1 that gives
minimum signal (destructive interference) at the listener’s loca-
tion? By what number must fmin,1 be multiplied to get (b) the sec-
ond lowest frequency fmin,2 that gives minimum signal and (c) the
third lowest frequency fmin,3 that gives minimum signal? (d) What is
the lowest frequency fmax,1 that gives maximum signal (constructive
interference) at the listener’s location? By what number must fmax,1
be multiplied to get (e) the second lowest frequency fmax,2 that
gives maximum signal and (f) the third lowest frequency fmax,3 that
gives maximum signal?
••18 In Fig. 17-34, sound waves A
and B, both of wavelength l, are ini-
tially in phase and traveling right-
ward, as indicated by the two rays.
Wave Ais reflected from four sur-
faces but ends up traveling in its orig-
inal direction.Wave Bends in that di-
rection after reflecting from two
surfaces. Let distance Lin the figure
be expressed as a multiple qof l:L
ILW
d d d
S1S2S3S4
P
Figure 17-36 Problem 20.
D
S1S2
Figure 17-35
Problems 19 and 105.
L
L
L
A
B
Figure 17-34 Problem 18.
Speakers
Listener
d1
d
2
Figure 17-37 Problem 21.
Source Detector
r
Figure 17-38 Problem 22.
S1P
S2
x
y
d
Figure 17-39 Problem 23.
508 CHAPTER 17 WAVES—II
1
100 500
r (m)
1000
(dB)
β
β
2
β
A
B
••34 Two atmospheric sound sources Aand Bemit isotropi-
cally at constant power. The sound levels bof their emissions are
plotted in Fig. 17-40 versus the radial distance rfrom the sources.
The vertical axis scale is set by b185.0 dB and b265.0 dB.
What are (a) the ratio of the larger power to the smaller power and
(b)the sound level difference at r10 m?
intensity of the waves 2.50 m from the source is 1.91 104W/m2.
Assuming that the energy of the waves is conserved, find the
power of the source.
•30 The source of a sound wave has a power of 1.00 mW. If it is a
point source, (a) what is the intensity 3.00 m away and (b) what is
the sound level in decibels at that distance?
•31 When you “crack” a knuckle, you suddenly widen
the knuckle cavity, allowing more volume for the synovial fluid in-
side it and causing a gas bubble suddenly to appear in the fluid.The
sudden production of the bubble, called “cavitation,” produces a
sound pulse—the cracking sound. Assume that the sound is trans-
mitted uniformly in all directions and that it fully passes from the
knuckle interior to the outside. If the pulse has a sound level of
62 dB at your ear, estimate the rate at which energy is produced by
the cavitation.
•32 Approximately a third of people with normal hearing
have ears that continuously emit a low-intensity sound outward
through the ear canal.A person with such spontaneous otoacoustic
emission is rarely aware of the sound, except perhaps in a noise-
free environment, but occasionally the emission is loud enough to
be heard by someone else nearby. In one observation, the sound
wave had a frequency of 1665 Hz and a pressure amplitude of
1.13 103Pa. What were (a) the displacement amplitude and
(b) the intensity of the wave emitted by the ear?
•33 Male Rana catesbeiana bullfrogs are known for their
loud mating call.The call is emitted not by the frog’s mouth but by
its eardrums, which lie on the surface of the head. And, surpris-
ingly, the sound has nothing to do with the frog’s inflated throat. If
the emitted sound has a frequency of 260 Hz and a sound level of
85 dB (near the eardrum), what is the amplitude of the eardrum’s
oscillation? The air density is 1.21 kg/m3.
the sources is d16.0 m. At any point Pon the xaxis, the wave
from S1and the wave from S2interfere. When Pis very far away
(x), what are (a) the phase difference between the arriving
waves from S1and S2and (b) the type of interference they pro-
duce? Now move point Palong the xaxis toward S1. (c) Does the
phase difference between the waves increase or decrease? At
what distance xdo the waves have a phase difference of (d)
0.50l, (e) 1.00l, and (f) 1.50l?
Module 17-4 Intensity and Sound Level
•24 Suppose that the sound level of a conversation is initially at
an angry 70 dB and then drops to a soothing 50 dB. Assuming that
the frequency of the sound is 500 Hz, determine the (a) initial and
(b) final sound intensities and the (c) initial and (d) final sound
wave amplitudes.
•25 A sound wave of frequency 300 Hz has an intensity of
1.00 mW/m2.What is the amplitude of the air oscillations caused by
this wave?
•26 A 1.0 W point source emits sound waves isotropically.
Assuming that the energy of the waves is conserved, find the inten-
sity (a) 1.0 m from the source and (b) 2.5 m from the source.
•27 A certain sound source is increased in sound
level by 30.0 dB. By what multiple is (a) its intensity increased and
(b) its pressure amplitude increased?
•28 Two sounds differ in sound level by 1.00 dB.What is the ratio
of the greater intensity to the smaller intensity?
•29 A point source emits sound waves isotropically. The
SSM
WWWSSM
Figure 17-40 Problem 34.
••35 A point source emits 30.0 W of sound isotropically. A small
microphone intercepts the sound in an area of 0.750 cm2, 200 m
from the source. Calculate (a) the sound intensity there and (b) the
power intercepted by the microphone.
••36 Party hearing. As the number of people at a party in-
creases, you must raise your voice for a listener to hear you against
the background noise of the other partygoers. However, once you
reach the level of yelling, the only way you can be heard is if you
move closer to your listener, into the listener’s “personal space.”
Model the situation by replacing you with an isotropic point source
of fixed power Pand replacing your listener with a point that ab-
sorbs part of your sound waves.These points are initially separated
by ri1.20 m. If the background noise increases by b5 dB, the
sound level at your listener must also increase. What separation rf
is then required?
•••37 A sound source sends a sinusoidal sound wave of angular
frequency 3000 rad/s and amplitude 12.0 nm through a tube of air.
The internal radius of the tube is 2.00 cm. (a) What is the average
rate at which energy (the sum of the kinetic and potential energies)
is transported to the opposite end of the tube? (b) If, simultane-
ously, an identical wave travels along an adjacent, identical tube,
what is the total average rate at which energy is transported to the
opposite ends of the two tubes by the waves? If, instead, those two
waves are sent along the same tube simultaneously, what is the to-
tal average rate at which they transport energy when their phase
difference is (c) 0, (d) 0.40prad, and (e) prad?
Module 17-5 Sources of Musical Sound
•38 The water level in a vertical glass tube 1.00 m long can be ad-
justed to any position in the tube.A tuning fork vibrating at 686 Hz
is held just over the open top end of the tube, to set up a standing
wave of sound in the air-filled top portion of the tube. (That air-
filled top portion acts as a tube with one end closed and the other
end open.) (a) For how many different positions of the water level
will sound from the fork set up resonance in the tube’s air-filled
portion? What are the (b) least and (c) second least water heights
in the tube for resonance to occur?
•39 (a) Find the speed of waves on a violin string of
mass 800 mg and length 22.0 cm if the fundamental frequency is
920 Hz. (b) What is the tension in the string? For the fundamental,
what is the wavelength of (c) the waves on the string and (d) the
sound waves emitted by the string?
ILWSSM
509
PROBLEMS
•40 Organ pipe A, with both ends open, has a fundamental
frequency of 300 Hz. The third harmonic of organ pipe B, with
one end open, has the same frequency as the second harmonic of
pipe A. How long are (a) pipe Aand (b) pipe B?
•41 A violin string 15.0 cm long and fixed at both ends oscillates
in its n1 mode. The speed of waves on the string is 250 m/s, and
the speed of sound in air is 348 m/s.What are the (a) frequency and
(b) wavelength of the emitted sound wave?
•42 A sound wave in a fluid medium is reflected at a barrier so
that a standing wave is formed. The distance between nodes is
3.8 cm, and the speed of propagation is 1500 m/s. Find the fre-
quency of the sound wave.
•43 In Fig. 17-41, Sis a small loudspeaker
driven by an audio oscillator with a frequency that
is varied from 1000 Hz to 2000 Hz, and Dis a cylin-
drical pipe with two open ends and a length of
45.7 cm.The speed of sound in the air-filled pipe is
344 m/s. (a) At how many frequencies does the
sound from the loudspeaker set up resonance in
the pipe? What are the (b) lowest and (c) second
lowest frequencies at which resonance occurs?
•44 The crest of a Parasaurolophus dinosaur skull is shaped
somewhat like a trombone and contains a nasal passage in the
form of a long, bent tube open at both ends. The dinosaur may
have used the passage to produce sound by setting up the funda-
mental mode in it. (a) If the nasal passage in a certain
Parasaurolophus fossil is 2.0 m long, what frequency would have
been produced? (b) If that dinosaur could be recreated (as in
Jurassic Park), would a person with a hearing range of 60 Hz to
20 kHz be able to hear that fundamental mode and, if so, would the
sound be high or low frequency? Fossil skulls that contain shorter
nasal passages are thought to be those of the female
Parasaurolophus. (c) Would that make the female’s fundamental
frequency higher or lower than the male’s?
•45 In pipe A, the ratio of a particular harmonic frequency to the
next lower harmonic frequency is 1.2. In pipe B, the ratio of a par-
ticular harmonic frequency to the next lower harmonic frequency
is 1.4. How many open ends are in (a) pipe Aand (b) pipe B?
••46 Pipe A, which is 1.20 m long and open at both ends,
oscillates at its third lowest harmonic frequency. It is filled with air
for which the speed of sound is 343 m/s. Pipe B, which is closed at
one end, oscillates at its second lowest harmonic frequency. This
frequency of Bhappens to match the frequency of A.An xaxis ex-
tends along the interior of B, with x0 at the closed end. (a) How
many nodes are along that axis? What are the (b) smallest and
(c) second smallest value of xlocating those nodes? (d) What is the
fundamental frequency of B?
••47 A well with vertical sides and water at the bottom resonates
at 7.00 Hz and at no lower frequency. The air-filled portion of the
well acts as a tube with one closed end (at the bottom) and one
open end (at the top).The air in the well has a density of 1.10 kg/m3
and a bulk modulus of 1.33 105Pa. How far down in the well is
the water surface?
••48 One of the harmonic frequencies of tube Awith two open
ends is 325 Hz. The next-highest harmonic frequency is 390 Hz.
(a) What harmonic frequency is next highest after the harmonic
frequency 195 Hz? (b) What is the number of this next-highest
harmonic? One of the harmonic frequencies of tube Bwith only
SSM
one open end is 1080 Hz. The next-highest harmonic frequency is
1320 Hz. (c) What harmonic frequency is next highest after the
harmonic frequency 600 Hz? (d) What is the number of this next-
highest harmonic?
••49 A violin string 30.0 cm long with linear density
0.650 g/m is placed near a loudspeaker that is fed by an audio oscil-
lator of variable frequency. It is found that the string is set into os-
cillation only at the frequencies 880 and 1320 Hz as the frequency
of the oscillator is varied over the range 500–1500 Hz. What is the
tension in the string?
••50 A tube 1.20 m long is closed at one end. A stretched wire
is placed near the open end. The wire is 0.330 m long and has a
mass of 9.60 g. It is fixed at both ends and oscillates in its funda-
mental mode. By resonance, it sets the air column in the tube into
oscillation at that column’s fundamental frequency. Find (a) that
frequency and (b) the tension in the wire.
Module 17-6 Beats
•51 The A string of a violin is a little too tightly stretched. Beats
at 4.00 per second are heard when the string is sounded together
with a tuning fork that is oscillating accurately at concert A
(440 Hz).What is the period of the violin string oscillation?
•52 A tuning fork of unknown frequency makes 3.00 beats per
second with a standard fork of frequency 384 Hz. The beat fre-
quency decreases when a small piece of wax is put on a prong of
the first fork.What is the frequency of this fork?
••53 Two identical piano wires have a fundamental
frequency of 600 Hz when kept under the same tension.What frac-
tional increase in the tension of one wire will lead to the occur-
rence of 6.0 beats/s when both wires oscillate simultaneously?
••54 You have five tuning forks that oscillate at close but differ-
ent resonant frequencies.What are the (a) maximum and (b) mini-
mum number of different beat frequencies you can produce by
sounding the forks two at a time, depending on how the resonant
frequencies differ?
Module 17-7 The Doppler Effect
•55 A whistle of frequency 540 Hz moves in a circle of radius
60.0 cm at an angular speed of 15.0 rad/s. What are the (a) lowest
and (b) highest frequencies heard by a listener a long distance
away, at rest with respect to the center of the circle?
•56 An ambulance with a siren emitting a whine at 1600 Hz over-
takes and passes a cyclist pedaling a bike at 2.44 m/s. After being
passed, the cyclist hears a frequency of 1590 Hz. How fast is the
ambulance moving?
•57 A state trooper chases a speeder along a straight road; both
vehicles move at 160 km/h.The siren on the trooper’s vehicle pro-
duces sound at a frequency of 500 Hz.What is the Doppler shift in
the frequency heard by the speeder?
••58 A sound source Aand a reflecting surface Bmove directly
toward each other. Relative to the air, the speed of source Ais
29.9 m/s, the speed of surface Bis 65.8 m/s, and the speed of sound
is 329 m/s. The source emits waves at frequency 1200 Hz as meas-
ured in the source frame. In the reflector frame, what are the
(a) frequency and (b) wavelength of the arriving sound waves? In
the source frame, what are the (c) frequency and (d) wavelength of
the sound waves reflected back to the source?
ILW
SSM
SSM
Figure 17-41
Problem 43.
S
D
510 CHAPTER 17 WAVES—II
Distal
sac Frontal
sac
S
permacet
i
sac
1.0 ms
Figure 17-44 Problem 73.
••59 In Fig. 17-42, a French submarine and a U.S. submarine
move toward each other during maneuvers in motionless water
in the North Atlantic. The French sub moves at speed vF
50.00 km/h, and the U.S. sub at vUS 70.00 km/h. The French sub
sends out a sonar signal (sound wave in water) at 1.000 103Hz.
Sonar waves travel at 5470 km/h. (a) What is the signal’s frequency
as detected by the U.S. sub? (b) What frequency is detected by the
French sub in the signal reflected back to it by the U.S. sub?
locomotive whistle emits sound at frequency 500.0 Hz. The air is
still. (a) What frequency does the uncle hear? (b) What frequency
does the girl hear? A wind begins to blow from the east at 10.00
m/s. (c) What frequency does the uncle now hear? (d) What fre-
quency does the girl now hear?
Module 17-8 Supersonic Speeds, Shock Waves
•68 The shock wave off the cockpit of the FA 18 in Fig. 17-24
has an angle of about 60. The airplane was traveling at about
1350 km/h when the photograph was taken. Approximately what
was the speed of sound at the airplane’s altitude?
••69 A jet plane passes over you at a height of 5000 m
and a speed of Mach 1.5. (a) Find the Mach cone angle (the sound
speed is 331 m/s). (b) How long after the jet passes directly over-
head does the shock wave reach you?
••70 A plane flies at 1.25 times the speed of sound.Its sonic boom
reaches a man on the ground 1.00 min after the plane passes di-
rectly overhead. What is the altitude of the plane? Assume the
speed of sound to be 330 m/s.
Additional Problems
71 At a distance of 10 km, a 100 Hz horn, assumed to be an
isotropic point source, is barely audible. At what distance would it
begin to cause pain?
72 A bullet is fired with a speed of 685 m/s. Find the angle made
by the shock cone with the line of motion of the bullet.
73 A sperm whale (Fig. 17-44a) vocalizes by producing a
series of clicks.Actually, the whale makes only a single sound near
the front of its head to start the series. Part of that sound then
emerges from the head into the water to become the first click of
the series. The rest of the sound travels backward through the
spermaceti sac (a body of fat), reflects from the frontal sac (an air
layer), and then travels forward through the spermaceti sac.When
it reaches the distal sac (another air layer) at the front of the head,
some of the sound escapes into the water to form the second click,
and the rest is sent back through the spermaceti sac (and ends up
forming later clicks).
Figure 17-44bshows a strip-chart recording of a series of clicks.
A unit time interval of 1.0 ms is indicated on the chart. Assuming
that the speed of sound in the spermaceti sac is 1372 m/s, find
the length of the spermaceti sac. From such a calculation, marine
scientists estimate the length of a whale from its click series.
SSM
French U.S.
vFvUS
Figure 17-42 Problem 59.
••60 A stationary motion detector sends sound waves of frequency
0.150 MHz toward a truck approaching at a speed of 45.0 m/s. What
is the frequency of the waves reflected back to the detector?
••61 A bat is flitting about in a cave, navigating via
ultrasonic bleeps.Assume that the sound emission frequency of the
bat is 39 000 Hz. During one fast swoop directly toward a flat wall
surface, the bat is moving at 0.025 times the speed of sound in air.
What frequency does the bat hear reflected off the wall?
••62 Figure 17-43 shows four tubes with lengths 1.0 m or 2.0 m,
with one or two open ends as drawn.The third harmonic is set up in
each tube, and some of the sound that escapes from them is detected
by detector D, which moves directly away from the tubes. In
terms of the speed of sound v,
what speed must the detector
have such that the detected
frequency of the sound from
(a) tube 1, (b) tube 2, (c) tube
3, and (d) tube 4 is equal to the
tube’s fundamental frequency?
••63 An acoustic burglar alarm consists of a source emitting
waves of frequency 28.0 kHz. What is the beat frequency between
the source waves and the waves reflected from an intruder walking
at an average speed of 0.950 m/s directly away from the alarm?
••64 A stationary detector measures the frequency of a sound
source that first moves at constant velocity directly toward the de-
tector and then (after passing the detector) directly away from it.
The emitted frequency is f. During the approach the detected fre-
quency is f
app and during the recession it is f
rec. If (f
app f
rec)/f
0.500, what is the ratio vs/vof the speed of the source to the speed
of sound?
•••65 A 2000 Hz siren and a civil defense official are both at
rest with respect to the ground. What frequency does the official
hear if the wind is blowing at 12 m/s (a) from source to official and
(b) from official to source?
•••66 Two trains are traveling toward each other at 30.5 m/s
relative to the ground. One train is blowing a whistle at 500 Hz.
(a) What frequency is heard on the other train in still air? (b) What
frequency is heard on the other train if the wind is blowing at
30.5 m/s toward the whistle and away from the listener? (c) What
frequency is heard if the wind direction is reversed?
•••67 A girl is sitting near the open window of a
train that is moving at a velocity of 10.00 m/s to the east.The girl’s
uncle stands near the tracks and watches the train move away.The
WWWSSM
ILW
D
1
2
3
4
Figure 17-43 Problem 62.
511
PROBLEMS
80 A detector initially moves at constant velocity directly
toward a stationary sound source and then (after passing it) di-
rectly from it. The emitted frequency is f. During the approach the
detected frequency is f
app and during the recession it is f
rec. If the
frequencies are related by (f
app f
rec)/f0.500, what is the ratio
vD/vof the speed of the detector to the speed of sound?
81 (a) If two sound waves, one in air and one in (fresh)
water, are equal in intensity and angular frequency, what is the
ratio of the pressure amplitude of the wave in water to that of
the wave in air? Assume the water and the air are at 20C. (See
Table 14-1.) (b) If the pressure amplitudes are equal instead, what
is the ratio of the intensities of the waves?
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74 The average density of Earth’s crust 10 km beneath the conti-
nents is 2.7 g/cm3. The speed of longitudinal seismic waves at that
depth, found by timing their arrival from distant earthquakes, is
5.4 km/s. Find the bulk modulus of Earth’s crust at that depth. For
comparison, the bulk modulus of steel is about 16 1010 Pa.
75 A certain loudspeaker system emits sound isotropically with
a frequency of 2000 Hz and an intensity of 0.960 mW/m2at a
distance of 6.10 m. Assume that there are no reflections. (a) What
is the intensity at 30.0 m? At 6.10 m, what are (b) the displacement
amplitude and (c) the pressure amplitude?
76 Find the ratios (greater to smaller) of the (a) intensities,
(b) pressure amplitudes, and (c) particle displacement amplitudes
for two sounds whose sound levels differ by 37 dB.
77 In Fig. 17-45, sound waves Aand B, both of wavelength l, are
initially in phase and traveling right-
ward, as indicated by the two rays.
Wave Ais reflected from four sur-
faces but ends up traveling in its
original direction. What multiple of
wavelength lis the smallest value of
distance Lin the figure that puts A
and Bexactly out of phase with each
other after the reflections?
78 A trumpet player on a moving
railroad flatcar moves toward a sec-
ond trumpet player standing alongside the track while both play
a 440 Hz note.The sound waves heard by a stationary observer be-
tween the two players have a beat frequency of 4.0 beats/s. What is
the flatcar’s speed?
79 In Fig. 17-46, sound of wavelength 0.850 m is emitted
isotropically by point source S. Sound ray 1 extends directly to
detector D, at distance L10.0 m. Sound ray 2 extends to Dvia a
reflection (effectively, a “bouncing”) of the sound at a flat surface.
That reflection occurs on a perpendicular bisector to the SD line,
at distance dfrom the line. Assume that the reflection shifts the
sound wave by 0.500l. For what least value of d(other than zero)
do the direct sound and the reflected sound arrive at D(a) exactly
out of phase and (b) exactly in phase?
82 A continuous sinusoidal longitudinal wave is sent along a very
long coiled spring from an attached oscillating source. The wave
travels in the negative direction of an xaxis; the source frequency
is 25 Hz; at any instant the distance between successive points of
maximum expansion in the spring is 24 cm; the maximum longitu-
dinal displacement of a spring particle is 0.30 cm; and the particle
at x0 has zero displacement at time t0. If the wave is written
in the form s(x,t)smcos(kx vt), what are (a) sm, (b) k, (c) v,
(d) the wave speed, and (e) the cor-
rect choice of sign in front of v?
83 Ultrasound, which consists
of sound waves with frequencies
above the human audible range, can
be used to produce an image of the
interior of a human body. Moreover,
ultrasound can be used to measure
the speed of the blood in the body; it
does so by comparing the frequency of the ultrasound sent into the
body with the frequency of the ultrasound reflected back to the
body’s surface by the blood. As the blood pulses, this detected fre-
quency varies.
Suppose that an ultrasound image of the arm of a patient shows
an artery that is angled at u20to the ultrasound’s line of travel
(Fig. 17-47). Suppose also that the frequency of the ultrasound
reflected by the blood in the artery is increased by a maximum of
5495 Hz from the original ultrasound frequency of 5.000 000 MHz.
(a) In Fig. 17-47, is the direction of the blood flow rightward or
leftward? (b) The speed of sound in the human arm is 1540 m/s.
What is the maximum speed of the blood? (Hint: The Doppler effect
is caused by the component of the blood’s velocity along the ultra-
sound’s direction of travel.) (c) If angle uwere greater, would the re-
flected frequency be greater or less?
84 The speed of sound in a certain metal is vm. One end of a
long pipe of that metal of length Lis struck a hard blow.
A listener at the other end hears two sounds, one from the wave
that travels along the pipe’s metal wall and the other from the
wave that travels through the air inside the pipe. (a) If vis the
speed of sound in air, what is the time interval tbetween the ar-
rivals of the two sounds at the listener’s ear? (b) If t1.00 s and
the metal is steel, what is the length L?
85 An avalanche of sand along some rare desert sand dunes
can produce a booming that is loud enough to be heard 10 km
away. The booming apparently results from a periodic oscillation
of the sliding layer of sand—the layer’s thickness expands and
contracts. If the emitted frequency is 90 Hz, what are (a) the period
of the thickness oscillation and (b) the wavelength of the sound?
86 A sound source moves along an xaxis, between detectors A
and B. The wavelength of the sound detected at Ais 0.500 that of
the sound detected at B. What is the ratio s/of the speed of the
source to the speed of sound?
87 A siren emitting a sound of frequency 1000 Hz moves
away from you toward the face of a cliff at a speed of 10 m/s. Take
the speed of sound in air as 330 m/s. (a) What is the frequency of
the sound you hear coming directly from the siren? (b) What is the
frequency of the sound you hear reflected off the cliff? (c) What is
the beat frequency between the two sounds? Is it perceptible (less
than 20 Hz)?
88 At a certain point, two waves produce pressure variations
given by p1pmsin vtand p2pmsin(vtf).At this point,
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SSM
S D
d
Ray 2
Ray 1
L
__
2
L
__
2
Figure 17-46 Problem 79.
L
L
A
B
Figure 17-45 Problem 77.
Incident
ultrasound
Artery
θ
Figure 17-47 Problem 83.
512 CHAPTER 17 WAVES—II
Figure 17-49 Problem 94.
Impact
Cylindrical
wave
Hemispherical wave
d
what is the ratio pr/pm, where pris the pressure amplitude of
the resultant wave, if fis (a) 0, (b) p/2, (c) p/3, and (d) p/4?
89 Two sound waves with an amplitude of 12 nm and a wave-
length of 35 cm travel in the same direction through a long tube,
with a phase difference of p/3 rad.What are the (a) amplitude and
(b) wavelength of the net sound wave produced by their interfer-
ence? If, instead, the sound waves travel through the tube in oppo-
site directions, what are the (c) amplitude and (d) wavelength of
the net wave?
90 A sinusoidal sound wave moves at 343 m/s through air in the
positive direction of an xaxis. At one instant during the oscilla-
tions, air molecule Ais at its maximum displacement in the nega-
tive direction of the axis while air molecule Bis at its equilibrium
position. The separation between those molecules is 15.0 cm, and
the molecules between Aand Bhave intermediate displacements
in the negative direction of the axis. (a) What is the frequency of
the sound wave?
In a similar arrangement but for a different sinusoidal sound
wave,at one instant air molecule Cis at its maximum displacement
in the positive direction while molecule Dis at its maximum
displacement in the negative direction.The separation between the
molecules is again 15.0 cm, and the molecules between Cand D
have intermediate displacements. (b) What is the frequency of the
sound wave?
91 Two identical tuning forks can oscillate at 440 Hz. A person is
located somewhere on the line between them. Calculate the beat
frequency as measured by this individual if (a) she is standing still
and the tuning forks move in the same direction along the line at
3.00 m/s, and (b) the tuning forks are stationary and the listener
moves along the line at 3.00 m/s.
92 You can estimate your distance from a lightning stroke by
counting the seconds between the flash you see and the thunder
you later hear. By what integer should you divide the number of
seconds to get the distance in kilometers?
93 Figure 17-48 shows an air-
filled, acoustic interferometer, used
to demonstrate the interference of
sound waves. Sound source Sis an
oscillating diaphragm; Dis a sound
detector, such as the ear or a micro-
phone. Path SBD can be varied in
length, but path SAD is fixed. At D,
the sound wave coming along path
SBD interferes with that coming along path SAD. In one demon-
stration, the sound intensity at Dhas a minimum value of
100 units at one position of the movable arm and continuously
climbs to a maximum value of 900 units when that arm is shifted
by 1.65 cm. Find (a) the frequency of the sound emitted by the
source and (b) the ratio of the amplitude at Dof the SAD wave
to that of the SBD wave. (c) How can it happen that these waves
have different amplitudes, considering that they originate at the
same source?
94 On July 10, 1996, a granite block broke away from a wall in
Yosemite Valley and, as it began to slide down the wall, was
launched into projectile motion. Seismic waves produced by its
impact with the ground triggered seismographs as far away as
200 km. Later measurements indicated that the block had a mass
between 7.3 107kg and 1.7 108kg and that it landed 500 m
vertically below the launch point and 30 m horizontally from it.
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(The launch angle is not known.) (a) Estimate the block’s kinetic
energy just before it landed.
Consider two types of seismic waves that spread from the im-
pact point—a hemispherical body wave traveled through the
ground in an expanding hemisphere and a cylindrical surface wave
traveled along the ground in an expanding shallow vertical cylin-
der (Fig. 17-49). Assume that the impact lasted 0.50 s, the vertical
cylinder had a depth dof 5.0 m, and each wave type received 20%
of the energy the block had just before impact. Neglecting any
mechanical energy loss the waves experienced as they traveled,
determine the intensities of (b) the body wave and (c) the surface
wave when they reached a seismograph 200 km away. (d) On the
basis of these results, which wave is more easily detected on a
distant seismograph?
S
A B
D
Figure 17-48 Problem 93.
95 The sound intensity is 0.0080 W/m2at a distance of 10 m
from an isotropic point source of sound. (a) What is the power of
the source? (b) What is the sound intensity 5.0 m from the source?
(c) What is the sound level 10 m from the source?
96 Four sound waves are to be sent through the same tube of air,
in the same direction:
s1(x,t)(9.00 nm) cos(2px700pt)
s2(x,t)(9.00 nm) cos(2px700pt0.7p)
s3(x,t)(9.00 nm) cos(2px700ptp)
s4(x,t)(9.00 nm) cos(2px700pt1.7p).
What is the amplitude of the resultant wave? (Hint: Use a phasor
diagram to simplify the problem.)
97 Straight line AB connects two point sources that are 5.00 m
apart, emit 300 Hz sound waves of the same amplitude, and emit
exactly out of phase. (a) What is the shortest distance between the
midpoint of AB and a point on AB where the interfering waves
cause maximum oscillation of the air molecules? What are the
(b) second and (c) third shortest distances?
98 A point source that is stationary on an xaxis emits a
sinusoidal sound wave at a frequency of 686 Hz and speed 343 m/s.
The wave travels radially outward from the source, causing air mol-
ecules to oscillate radially inward and outward. Let us define a
wavefront as a line that connects points where the air molecules
have the maximum, radially outward displacement. At any given
instant, the wavefronts are concentric circles that are centered on
the source. (a) Along x, what is the adjacent wavefront separation?
Next, the source moves along xat a speed of 110 m/s. Along x,
what are the wavefront separations (b) in front of and (c) behind
the source?
99 You are standing at a distance Dfrom an isotropic point source
of sound. You walk 50.0 m toward the source and observe that the
intensity of the sound has doubled. Calculate the distance D.
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513
PROBLEMS
100 Pipe Ahas only one open end; pipe Bis four times as long
and has two open ends. Of the lowest 10 harmonic numbers nBof
pipe B, what are the (a) smallest, (b) second smallest, and (c) third
smallest values at which a harmonic frequency of Bmatches one of
the harmonic frequencies of A?
101 A pipe 0.60 m long and closed at one end is filled with an
unknown gas. The third lowest harmonic frequency for the pipe is
750 Hz. (a) What is the speed of sound in the unknown gas?
(b) What is the fundamental frequency for this pipe when it is filled
with the unknown gas?
102 A sound wave travels out uniformly in all directions from a
point source. (a) Justify the following expression for the displace-
ment sof the transmitting medium at any distance rfrom the source:
where bis a constant. Consider the speed, direction of propaga-
tion, periodicity, and intensity of the wave. (b) What is the dimen-
sion of the constant b?
103 A police car is chasing a speeding Porsche 911. Assume that
the Porsche’s maximum speed is 80.0 m/s and the police car’s is 54.0
m/s.At the moment both cars reach their maximum speed, what fre-
quency will the Porsche driver hear if the frequency of the police
car’s siren is 440 Hz? Take the speed of sound in air to be 340 m/s.
104 Suppose a spherical loudspeaker emits sound isotropically at
10 W into a room with completely absorbent walls, floor, and ceiling
(an anechoic chamber). (a) What is the intensity of the sound at
distance d3.0 m from the center of the source? (b) What is the
ratio of the wave amplitude at d4.0 m to that at d3.0 m?
105 In Fig. 17-35, S1and S2are two isotropic point sources of
sound. They emit waves in phase at wavelength 0.50 m; they are
separated by D1.60 m. If we move a sound detector along a
large circle centered at the midpoint between the sources, at how
many points do waves arrive at the detector (a) exactly in phase
and (b) exactly out of phase?
106 Figure 17-50 shows a transmitter and receiver of waves con-
tained in a single instrument. It is used to measure the speed uof a
target object (idealized as a flat plate) that is moving directly to-
ward the unit, by analyzing the waves reflected from the target.
What is uif the emitted frequency is 18.0 kHz and the detected fre-
quency (of the returning waves) is 22.2 kHz?
sb
r sin k(rvt),
plunger Pis provided at the other end of the tube, and the tube is
filled with a gas.The rod is made to oscillate longitudinally at fre-
quency fto produce sound waves inside the gas, and the location
of the plunger is adjusted until a standing sound wave pattern is
set up inside the tube. Once the standing wave is set up, the mo-
tion of the gas molecules causes the cork filings to collect in a
pattern of ridges at the displacement nodes. If f4.46 103Hz
and the separation between ridges is 9.20 cm, what is the speed of
sound in the gas?
108 A source S and a detector D of radio waves are a distance d
apart on level ground (Fig. 17-52). Radio waves of wavelength l
reach D either along a straight path or by reflecting (bouncing)
from a certain layer in the atmosphere. When the layer is at height
H, the two waves reaching D are exactly in phase. If the layer grad-
ually rises, the phase difference between the two waves gradually
shifts, until they are exactly out of phase when the layer is at height
Hh. Express lin terms of d,h, and H.
Figure 17-50 Problem 106.
Targe
t
f
r
f
s
u
Figure 17-51 Problem 107.
PD
R
d
110 A person on a railroad car blows a trumpet note at 440 Hz.
The car is moving toward a wall at 20.0 m/s. Find the sound fre-
quency (a) at the wall and (b) reflected back to the trumpeter.
111 A listener at rest (with respect to the air and the ground)
hears a signal of frequency f1from a source moving toward him
with a velocity of 15 m/s, due east. If the listener then moves toward
the approaching source with a velocity of 25 m/s, due west, he hears
a frequency f2that differs from f1by 37 Hz.What is the frequency of
the source? (Take the speed of sound in air to be 340 m/s.)
107 Kundt’s method for measuring the speed of sound. In Fig.
17-51, a rod Ris clamped at its center; a disk Dat its end projects
into a glass tube that has cork filings spread over its interior. A
Figure 17-52 Problem 108.
SD
H
d/2 d/2
h
109 In Fig. 17-53, a point source Sof sound waves lies near a
reflecting wall AB. A sound detector Dintercepts sound ray R1
traveling directly from S. It also intercepts sound ray R2that re-
flects from the wall such that the angle of incidence uiis equal to
the angle of reflection ur. Assume that the reflection of sound by
the wall causes a phase shift of 0.500l. If the distances are d1
2.50 m, d220.0 m, and d312.5 m, what are the (a) lowest and
(b) second lowest frequency at which R1and R2are in phase at D?
Figure 17-53 Problem 109.
D
A
B
S
d2
d3
R1
R2
ur
ui
d1
